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feat: add solutions to lc problem: No.2077
No.2077.Paths in Maze That Lead to Same Room
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‎solution/2000-2099/2077.Paths in Maze That Lead to Same Room/README.md

+110-1
Original file line numberDiff line numberDiff line change
@@ -56,22 +56,131 @@ There are no cycles of length 3.
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<!-- 这里可写通用的实现逻辑 -->
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**方法一：哈希表**
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长度为 `3` 的环，由三个顶点、三条边组成。我们假设三个顶点分别为 `a`, `b`, `c`。
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那么一定存在 `c <=> a`，`c <=> b` 以及 `a <=> b`，即环中的每个点都与其他两点直接相连。我们可以用哈希表来存放每个点的相邻点。
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由于环的长度为 `3`，每个相同的环会被重复统计 `3` 次，因此答案需除以 `3`。
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时间复杂度 $O(n^2)$，空间复杂度 $O(n)$。
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def numberOfPaths(self, n: int, corridors: List[List[int]]) -> int:
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g = defaultdict(set)
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for a, b in corridors:
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g[a].add(b)
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g[b].add(a)
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ans = 0
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for i in range(1, n + 1):
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for j, k in combinations(g[i], 2):
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if j in g[k]:
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ans += 1
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return ans // 3
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public int numberOfPaths(int n, int[][] corridors) {
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Set<Integer>[] g = new Set[n + 1];
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for (int i = 0; i <= n; ++i) {
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g[i] = new HashSet<>();
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}
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for (var c : corridors) {
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int a = c[0], b = c[1];
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g[a].add(b);
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g[b].add(a);
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}
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int ans = 0;
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for (int c = 1; c <= n; ++c) {
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var nxt = new ArrayList<>(g[c]);
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int m = nxt.size();
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for (int i = 0; i < m; ++i) {
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for (int j = i + 1; j < m; ++j) {
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int a = nxt.get(i), b = nxt.get(j);
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if (g[b].contains(a)) {
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++ans;
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}
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}
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}
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}
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return ans / 3;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int numberOfPaths(int n, vector<vector<int>>& corridors) {
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vector<unordered_set<int>> g(n + 1);
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for (auto& c : corridors) {
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int a = c[0], b = c[1];
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g[a].insert(b);
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g[b].insert(a);
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}
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int ans = 0;
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for (int c = 1; c <= n; ++c) {
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vector<int> nxt;
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nxt.assign(g[c].begin(), g[c].end());
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int m = nxt.size();
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for (int i = 0; i < m; ++i) {
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for (int j = i + 1; j < m; ++j) {
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int a = nxt[i], b = nxt[j];
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ans += g[b].count(a);
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}
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}
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}
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return ans / 3;
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}
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};
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```
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### **Go**
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```go
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func numberOfPaths(n int, corridors [][]int) int {
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g := make([]map[int]bool, n+1)
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for i := range g {
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g[i] = make(map[int]bool)
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}
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for _, c := range corridors {
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a, b := c[0], c[1]
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g[a][b] = true
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g[b][a] = true
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}
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ans := 0
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for c := 1; c <= n; c++ {
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nxt := []int{}
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for v := range g[c] {
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nxt = append(nxt, v)
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}
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m := len(nxt)
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for i := 0; i < m; i++ {
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for j := i + 1; j < m; j++ {
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a, b := nxt[i], nxt[j]
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if g[b][a] {
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ans++
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}
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}
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}
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}
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return ans / 3
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}
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```
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### **...**

‎solution/2000-2099/2077.Paths in Maze That Lead to Same Room/README_EN.md

+100-1
Original file line numberDiff line numberDiff line change
@@ -57,13 +57,112 @@ There are no cycles of length 3.
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### **Python3**
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```python
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class Solution:
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def numberOfPaths(self, n: int, corridors: List[List[int]]) -> int:
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g = defaultdict(set)
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for a, b in corridors:
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g[a].add(b)
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g[b].add(a)
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ans = 0
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for i in range(1, n + 1):
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for j, k in combinations(g[i], 2):
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if j in g[k]:
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ans += 1
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return ans // 3
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```
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### **Java**
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```java
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class Solution {
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public int numberOfPaths(int n, int[][] corridors) {
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Set<Integer>[] g = new Set[n + 1];
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for (int i = 0; i <= n; ++i) {
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g[i] = new HashSet<>();
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}
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for (var c : corridors) {
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int a = c[0], b = c[1];
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g[a].add(b);
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g[b].add(a);
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}
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int ans = 0;
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for (int c = 1; c <= n; ++c) {
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var nxt = new ArrayList<>(g[c]);
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int m = nxt.size();
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for (int i = 0; i < m; ++i) {
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for (int j = i + 1; j < m; ++j) {
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int a = nxt.get(i), b = nxt.get(j);
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if (g[b].contains(a)) {
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++ans;
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}
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}
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}
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}
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return ans / 3;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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int numberOfPaths(int n, vector<vector<int>>& corridors) {
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vector<unordered_set<int>> g(n + 1);
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for (auto& c : corridors) {
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int a = c[0], b = c[1];
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g[a].insert(b);
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g[b].insert(a);
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}
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int ans = 0;
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for (int c = 1; c <= n; ++c) {
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vector<int> nxt;
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nxt.assign(g[c].begin(), g[c].end());
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int m = nxt.size();
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for (int i = 0; i < m; ++i) {
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for (int j = i + 1; j < m; ++j) {
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int a = nxt[i], b = nxt[j];
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ans += g[b].count(a);
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}
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}
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}
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return ans / 3;
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}
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};
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```
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### **Go**
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```go
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func numberOfPaths(n int, corridors [][]int) int {
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g := make([]map[int]bool, n+1)
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for i := range g {
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g[i] = make(map[int]bool)
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}
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for _, c := range corridors {
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a, b := c[0], c[1]
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g[a][b] = true
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g[b][a] = true
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}
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ans := 0
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for c := 1; c <= n; c++ {
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nxt := []int{}
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for v := range g[c] {
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nxt = append(nxt, v)
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}
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m := len(nxt)
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for i := 0; i < m; i++ {
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for j := i + 1; j < m; j++ {
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a, b := nxt[i], nxt[j]
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if g[b][a] {
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ans++
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}
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}
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}
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}
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return ans / 3
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}
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```
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### **...**

‎solution/2000-2099/2077.Paths in Maze That Lead to Same Room/Solution.cpp

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Original file line numberDiff line numberDiff line change
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class Solution {
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public:
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int numberOfPaths(int n, vector<vector<int>>& corridors) {
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vector<unordered_set<int>> g(n + 1);
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for (auto& c : corridors) {
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int a = c[0], b = c[1];
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g[a].insert(b);
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g[b].insert(a);
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}
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int ans = 0;
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for (int c = 1; c <= n; ++c) {
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vector<int> nxt;
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nxt.assign(g[c].begin(), g[c].end());
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int m = nxt.size();
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for (int i = 0; i < m; ++i) {
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for (int j = i + 1; j < m; ++j) {
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int a = nxt[i], b = nxt[j];
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ans += g[b].count(a);
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}
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}
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}
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return ans / 3;
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}
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};

‎solution/2000-2099/2077.Paths in Maze That Lead to Same Room/Solution.go

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func numberOfPaths(n int, corridors [][]int) int {
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g := make([]map[int]bool, n+1)
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for i := range g {
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g[i] = make(map[int]bool)
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}
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for _, c := range corridors {
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a, b := c[0], c[1]
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g[a][b] = true
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g[b][a] = true
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}
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ans := 0
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for c := 1; c <= n; c++ {
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nxt := []int{}
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for v := range g[c] {
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nxt = append(nxt, v)
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}
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m := len(nxt)
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for i := 0; i < m; i++ {
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for j := i + 1; j < m; j++ {
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a, b := nxt[i], nxt[j]
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if g[b][a] {
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ans++
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}
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}
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}
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}
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return ans / 3
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}

‎solution/2000-2099/2077.Paths in Maze That Lead to Same Room/Solution.java

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@@ -0,0 +1,27 @@
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class Solution {
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public int numberOfPaths(int n, int[][] corridors) {
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Set<Integer>[] g = new Set[n + 1];
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for (int i = 0; i <= n; ++i) {
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g[i] = new HashSet<>();
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}
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for (var c : corridors) {
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int a = c[0], b = c[1];
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g[a].add(b);
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g[b].add(a);
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}
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int ans = 0;
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for (int c = 1; c <= n; ++c) {
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var nxt = new ArrayList<>(g[c]);
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int m = nxt.size();
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for (int i = 0; i < m; ++i) {
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for (int j = i + 1; j < m; ++j) {
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int a = nxt.get(i), b = nxt.get(j);
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if (g[b].contains(a)) {
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++ans;
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}
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}
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}
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}
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return ans / 3;
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}
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}

‎solution/2000-2099/2077.Paths in Maze That Lead to Same Room/Solution.py

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Original file line numberDiff line numberDiff line change
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class Solution:
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def numberOfPaths(self, n: int, corridors: List[List[int]]) -> int:
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g = defaultdict(set)
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for a, b in corridors:
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g[a].add(b)
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g[b].add(a)
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ans = 0
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for i in range(1, n + 1):
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for j, k in combinations(g[i], 2):
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if j in g[k]:
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ans += 1
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return ans // 3

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