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Oct 8, 2024
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feat: update solutions to lc problems: No.1436,2161 #3611

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76 changes: 21 additions & 55 deletions solution/1400-1499/1436.Destination City/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -30,7 +30,7 @@ tags:

<pre>
<strong>输入：</strong>paths = [["London","New York"],["New York","Lima"],["Lima","Sao Paulo"]]
<strong>输出：</strong>"Sao Paulo"
<strong>输出：</strong>"Sao Paulo"
<strong>解释：</strong>从 "London" 出发，最后抵达终点站 "Sao Paulo" 。本次旅行的路线是 "London" -&gt; "New York" -&gt; "Lima" -&gt; "Sao Paulo" 。
</pre>

Expand Down Expand Up @@ -74,9 +74,9 @@ tags:

### 方法一：哈希表

将所有起点存入哈希表中，然后遍历所有终点，找出没出现在哈希表中的终点，即为答案。
根据题目描述，终点一定不会出现在所有 $\textit{cityA}$ 中，因此，我们可以先遍历一遍 $\textit{paths}$，将所有 $\textit{cityA}$ 放入一个集合 $\textit{s}$ 中，然后再遍历一遍 $\textit{paths}$，找到不在 $\textit{s}$ 中的 $\textit{cityB}$ 即可。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是线路数。
时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为 $\textit{paths}$ 的长度。

<!-- tabs:start -->

Expand All @@ -98,12 +98,12 @@ class Solution {
for (var p : paths) {
s.add(p.get(0));
}
for (var p : paths) {
if (!s.contains(p.get(1))) {
return p.get(1);
for (int i = 0;; ++i) {
var b = paths.get(i).get(1);
if (!s.contains(b)) {
return b;
}
}
return "";
}
}
```
Expand All @@ -118,12 +118,12 @@ public:
for (auto& p : paths) {
s.insert(p[0]);
}
for (auto& p : paths) {
if (!s.count(p[1])) {
return p[1];
for (int i = 0;; ++i) {
auto b = paths[i][1];
if (!s.contains(b)) {
return b;
}
}
return "";
}
};
```
Expand All @@ -149,29 +149,23 @@ func destCity(paths [][]string) string {

```ts
function destCity(paths: string[][]): string {
const set = new Set(paths.map(([a]) => a));
for (const [_, b] of paths) {
if (!set.has(b)) {
return b;
}
}
return '';
const s = new Set<string>(paths.map(([a, _]) => a));
return paths.find(([_, b]) => !s.has(b))![1];
}
```

#### Rust

```rust
use std::collections::HashSet;

impl Solution {
pub fn dest_city(paths: Vec<Vec<String>>) -> String {
let set = paths.iter().map(|v| &v[0]).collect::<HashSet<&String>>();
for path in paths.iter() {
if !set.contains(&path[1]) {
return path[1].clone();
}
}
String::new()
let s = paths
.iter()
.map(|p| p[0].clone())
.collect::<HashSet<String>>();
paths.into_iter().find(|p| !s.contains(&p[1])).unwrap()[1].clone()
}
}
```
Expand All @@ -184,39 +178,11 @@ impl Solution {
* @return {string}
*/
var destCity = function (paths) {
const s = new Set();
for (const [a, _] of paths) {
s.add(a);
}
for (const [_, b] of paths) {
if (!s.has(b)) {
return b;
}
}
return '';
const s = new Set(paths.map(([a, _]) => a));
return paths.find(([_, b]) => !s.has(b))[1];
};
```

#### C

```c
char* destCity(char*** paths, int pathsSize, int* pathsColSize) {
for (int i = 0; i < pathsSize; i++) {
int flag = 1;
for (int j = 0; j < pathsSize; j++) {
if (strcmp(paths[i][1], paths[j][0]) == 0) {
flag = 0;
break;
}
}
if (flag) {
return paths[i][1];
}
}
return NULL;
}
```

<!-- tabs:end -->

<!-- solution:end -->
Expand Down
78 changes: 24 additions & 54 deletions solution/1400-1499/1436.Destination City/README_EN.md
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Original file line number Diff line number Diff line change
Expand Up @@ -29,7 +29,7 @@ tags:

<pre>
<strong>Input:</strong> paths = [[&quot;London&quot;,&quot;New York&quot;],[&quot;New York&quot;,&quot;Lima&quot;],[&quot;Lima&quot;,&quot;Sao Paulo&quot;]]
<strong>Output:</strong> &quot;Sao Paulo&quot;
<strong>Output:</strong> &quot;Sao Paulo&quot;
<strong>Explanation:</strong> Starting at &quot;London&quot; city you will reach &quot;Sao Paulo&quot; city which is the destination city. Your trip consist of: &quot;London&quot; -&gt; &quot;New York&quot; -&gt; &quot;Lima&quot; -&gt; &quot;Sao Paulo&quot;.
</pre>

Expand Down Expand Up @@ -70,7 +70,11 @@ Clearly the destination city is &quot;A&quot;.

<!-- solution:start -->

### Solution 1
### Solution 1: Hash Table

According to the problem description, the destination city will not appear in any of the $\textit{cityA}$. Therefore, we can first traverse the $\textit{paths}$ and put all $\textit{cityA}$ into a set $\textit{s}$. Then, we traverse the $\textit{paths}$ again to find the $\textit{cityB}$ that is not in $\textit{s}$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of $\textit{paths}$.

<!-- tabs:start -->

Expand All @@ -92,12 +96,12 @@ class Solution {
for (var p : paths) {
s.add(p.get(0));
}
for (var p : paths) {
if (!s.contains(p.get(1))) {
return p.get(1);
for (int i = 0;; ++i) {
var b = paths.get(i).get(1);
if (!s.contains(b)) {
return b;
}
}
return "";
}
}
```
Expand All @@ -112,12 +116,12 @@ public:
for (auto& p : paths) {
s.insert(p[0]);
}
for (auto& p : paths) {
if (!s.count(p[1])) {
return p[1];
for (int i = 0;; ++i) {
auto b = paths[i][1];
if (!s.contains(b)) {
return b;
}
}
return "";
}
};
```
Expand All @@ -143,29 +147,23 @@ func destCity(paths [][]string) string {

```ts
function destCity(paths: string[][]): string {
const set = new Set(paths.map(([a]) => a));
for (const [_, b] of paths) {
if (!set.has(b)) {
return b;
}
}
return '';
const s = new Set<string>(paths.map(([a, _]) => a));
return paths.find(([_, b]) => !s.has(b))![1];
}
```

#### Rust

```rust
use std::collections::HashSet;

impl Solution {
pub fn dest_city(paths: Vec<Vec<String>>) -> String {
let set = paths.iter().map(|v| &v[0]).collect::<HashSet<&String>>();
for path in paths.iter() {
if !set.contains(&path[1]) {
return path[1].clone();
}
}
String::new()
let s = paths
.iter()
.map(|p| p[0].clone())
.collect::<HashSet<String>>();
paths.into_iter().find(|p| !s.contains(&p[1])).unwrap()[1].clone()
}
}
```
Expand All @@ -178,39 +176,11 @@ impl Solution {
* @return {string}
*/
var destCity = function (paths) {
const s = new Set();
for (const [a, _] of paths) {
s.add(a);
}
for (const [_, b] of paths) {
if (!s.has(b)) {
return b;
}
}
return '';
const s = new Set(paths.map(([a, _]) => a));
return paths.find(([_, b]) => !s.has(b))[1];
};
```

#### C

```c
char* destCity(char*** paths, int pathsSize, int* pathsColSize) {
for (int i = 0; i < pathsSize; i++) {
int flag = 1;
for (int j = 0; j < pathsSize; j++) {
if (strcmp(paths[i][1], paths[j][0]) == 0) {
flag = 0;
break;
}
}
if (flag) {
return paths[i][1];
}
}
return NULL;
}
```

<!-- tabs:end -->

<!-- solution:end -->
Expand Down
15 changes: 0 additions & 15 deletions solution/1400-1499/1436.Destination City/Solution.c
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This file was deleted.

10 changes: 5 additions & 5 deletions solution/1400-1499/1436.Destination City/Solution.cpp
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Original file line number Diff line number Diff line change
Expand Up @@ -5,11 +5,11 @@ class Solution {
for (auto& p : paths) {
s.insert(p[0]);
}
for (auto& p : paths) {
if (!s.count(p[1])) {
return p[1];
for (int i = 0;; ++i) {
auto b = paths[i][1];
if (!s.contains(b)) {
return b;
}
}
return "";
}
};
};
10 changes: 5 additions & 5 deletions solution/1400-1499/1436.Destination City/Solution.java
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Original file line number Diff line number Diff line change
Expand Up @@ -4,11 +4,11 @@ public String destCity(List<List<String>> paths) {
for (var p : paths) {
s.add(p.get(0));
}
for (var p : paths) {
if (!s.contains(p.get(1))) {
return p.get(1);
for (int i = 0;; ++i) {
var b = paths.get(i).get(1);
if (!s.contains(b)) {
return b;
}
}
return "";
}
}
}
12 changes: 2 additions & 10 deletions solution/1400-1499/1436.Destination City/Solution.js
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Original file line number Diff line number Diff line change
Expand Up @@ -3,14 +3,6 @@
* @return {string}
*/
var destCity = function (paths) {
const s = new Set();
for (const [a, _] of paths) {
s.add(a);
}
for (const [_, b] of paths) {
if (!s.has(b)) {
return b;
}
}
return '';
const s = new Set(paths.map(([a, _]) => a));
return paths.find(([_, b]) => !s.has(b))[1];
};
13 changes: 6 additions & 7 deletions solution/1400-1499/1436.Destination City/Solution.rs
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Original file line number Diff line number Diff line change
@@ -1,12 +1,11 @@
use std::collections::HashSet;

impl Solution {
pub fn dest_city(paths: Vec<Vec<String>>) -> String {
let set = paths.iter().map(|v| &v[0]).collect::<HashSet<&String>>();
for path in paths.iter() {
if !set.contains(&path[1]) {
return path[1].clone();
}
}
String::new()
let s = paths
.iter()
.map(|p| p[0].clone())
.collect::<HashSet<String>>();
paths.into_iter().find(|p| !s.contains(&p[1])).unwrap()[1].clone()
}
}
9 changes: 2 additions & 7 deletions solution/1400-1499/1436.Destination City/Solution.ts
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Original file line number Diff line number Diff line change
@@ -1,9 +1,4 @@
function destCity(paths: string[][]): string {
const set = new Set(paths.map(([a]) => a));
for (const [_, b] of paths) {
if (!set.has(b)) {
return b;
}
}
return '';
const s = new Set<string>(paths.map(([a, _]) => a));
return paths.find(([_, b]) => !s.has(b))![1];
}
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