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feat: update solutions to lc problems: No.1436,2161 (#3611)
* No.1436.Destination City * No.2161.Partition Array According to Given Pivot
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‎solution/1400-1499/1436.Destination City/README.md

Lines changed: 21 additions & 55 deletions
Original file line numberDiff line numberDiff line change
@@ -30,7 +30,7 @@ tags:
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<pre>
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<strong>输入：</strong>paths = [["London","New York"],["New York","Lima"],["Lima","Sao Paulo"]]
33-
<strong>输出：</strong>"Sao Paulo"
33+
<strong>输出：</strong>"Sao Paulo"
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<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" 。
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</pre>
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@@ -74,9 +74,9 @@ tags:
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### 方法一：哈希表
7676

77-
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。
77+
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。
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79-
时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 是线路数。
79+
时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为 $\textit{paths}$ 的长度。
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<!-- tabs:start -->
8282

@@ -98,12 +98,12 @@ class Solution {
9898
for (var p : paths) {
9999
s.add(p.get(0));
100100
}
101-
for (var p : paths) {
102-
if (!s.contains(p.get(1))) {
103-
return p.get(1);
101+
for (int i = 0;; ++i) {
102+
var b = paths.get(i).get(1);
103+
if (!s.contains(b)) {
104+
return b;
104105
}
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}
106-
return "";
107107
}
108108
}
109109
```
@@ -118,12 +118,12 @@ public:
118118
for (auto& p : paths) {
119119
s.insert(p[0]);
120120
}
121-
for (auto& p : paths) {
122-
if (!s.count(p[1])) {
123-
return p[1];
121+
for (int i = 0;; ++i) {
122+
auto b = paths[i][1];
123+
if (!s.contains(b)) {
124+
return b;
124125
}
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}
126-
return "";
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}
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};
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```
@@ -149,29 +149,23 @@ func destCity(paths [][]string) string {
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150150
```ts
151151
function destCity(paths: string[][]): string {
152-
const set = new Set(paths.map(([a]) => a));
153-
for (const [_, b] of paths) {
154-
if (!set.has(b)) {
155-
return b;
156-
}
157-
}
158-
return '';
152+
const s = new Set<string>(paths.map(([a, _]) => a));
153+
return paths.find(([_, b]) => !s.has(b))![1];
159154
}
160155
```
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#### Rust
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164159
```rust
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use std::collections::HashSet;
161+
166162
impl Solution {
167163
pub fn dest_city(paths: Vec<Vec<String>>) -> String {
168-
let set = paths.iter().map(|v| &v[0]).collect::<HashSet<&String>>();
169-
for path in paths.iter() {
170-
if !set.contains(&path[1]) {
171-
return path[1].clone();
172-
}
173-
}
174-
String::new()
164+
let s = paths
165+
.iter()
166+
.map(|p| p[0].clone())
167+
.collect::<HashSet<String>>();
168+
paths.into_iter().find(|p| !s.contains(&p[1])).unwrap()[1].clone()
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}
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}
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```
@@ -184,39 +178,11 @@ impl Solution {
184178
* @return {string}
185179
*/
186180
var destCity = function (paths) {
187-
const s = new Set();
188-
for (const [a, _] of paths) {
189-
s.add(a);
190-
}
191-
for (const [_, b] of paths) {
192-
if (!s.has(b)) {
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return b;
194-
}
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}
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return '';
181+
const s = new Set(paths.map(([a, _]) => a));
182+
return paths.find(([_, b]) => !s.has(b))[1];
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};
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```
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200-
#### C
201-
202-
```c
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char* destCity(char*** paths, int pathsSize, int* pathsColSize) {
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for (int i = 0; i < pathsSize; i++) {
205-
int flag = 1;
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for (int j = 0; j < pathsSize; j++) {
207-
if (strcmp(paths[i][1], paths[j][0]) == 0) {
208-
flag = 0;
209-
break;
210-
}
211-
}
212-
if (flag) {
213-
return paths[i][1];
214-
}
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}
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return NULL;
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}
218-
```
219-
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/1400-1499/1436.Destination City/README_EN.md

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@@ -29,7 +29,7 @@ tags:
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<pre>
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<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;]]
32-
<strong>Output:</strong> &quot;Sao Paulo&quot;
32+
<strong>Output:</strong> &quot;Sao Paulo&quot;
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<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;.
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</pre>
3535

@@ -70,7 +70,11 @@ Clearly the destination city is &quot;A&quot;.
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<!-- solution:start -->
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73-
### Solution 1
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### Solution 1: Hash Table
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75+
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 $\textit{s}$.
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77+
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of $\textit{paths}$.
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<!-- tabs:start -->
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@@ -92,12 +96,12 @@ class Solution {
9296
for (var p : paths) {
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s.add(p.get(0));
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}
95-
for (var p : paths) {
96-
if (!s.contains(p.get(1))) {
97-
return p.get(1);
99+
for (int i = 0;; ++i) {
100+
var b = paths.get(i).get(1);
101+
if (!s.contains(b)) {
102+
return b;
98103
}
99104
}
100-
return "";
101105
}
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}
103107
```
@@ -112,12 +116,12 @@ public:
112116
for (auto& p : paths) {
113117
s.insert(p[0]);
114118
}
115-
for (auto& p : paths) {
116-
if (!s.count(p[1])) {
117-
return p[1];
119+
for (int i = 0;; ++i) {
120+
auto b = paths[i][1];
121+
if (!s.contains(b)) {
122+
return b;
118123
}
119124
}
120-
return "";
121125
}
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};
123127
```
@@ -143,29 +147,23 @@ func destCity(paths [][]string) string {
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144148
```ts
145149
function destCity(paths: string[][]): string {
146-
const set = new Set(paths.map(([a]) => a));
147-
for (const [_, b] of paths) {
148-
if (!set.has(b)) {
149-
return b;
150-
}
151-
}
152-
return '';
150+
const s = new Set<string>(paths.map(([a, _]) => a));
151+
return paths.find(([_, b]) => !s.has(b))![1];
153152
}
154153
```
155154

156155
#### Rust
157156

158157
```rust
159158
use std::collections::HashSet;
159+
160160
impl Solution {
161161
pub fn dest_city(paths: Vec<Vec<String>>) -> String {
162-
let set = paths.iter().map(|v| &v[0]).collect::<HashSet<&String>>();
163-
for path in paths.iter() {
164-
if !set.contains(&path[1]) {
165-
return path[1].clone();
166-
}
167-
}
168-
String::new()
162+
let s = paths
163+
.iter()
164+
.map(|p| p[0].clone())
165+
.collect::<HashSet<String>>();
166+
paths.into_iter().find(|p| !s.contains(&p[1])).unwrap()[1].clone()
169167
}
170168
}
171169
```
@@ -178,39 +176,11 @@ impl Solution {
178176
* @return {string}
179177
*/
180178
var destCity = function (paths) {
181-
const s = new Set();
182-
for (const [a, _] of paths) {
183-
s.add(a);
184-
}
185-
for (const [_, b] of paths) {
186-
if (!s.has(b)) {
187-
return b;
188-
}
189-
}
190-
return '';
179+
const s = new Set(paths.map(([a, _]) => a));
180+
return paths.find(([_, b]) => !s.has(b))[1];
191181
};
192182
```
193183

194-
#### C
195-
196-
```c
197-
char* destCity(char*** paths, int pathsSize, int* pathsColSize) {
198-
for (int i = 0; i < pathsSize; i++) {
199-
int flag = 1;
200-
for (int j = 0; j < pathsSize; j++) {
201-
if (strcmp(paths[i][1], paths[j][0]) == 0) {
202-
flag = 0;
203-
break;
204-
}
205-
}
206-
if (flag) {
207-
return paths[i][1];
208-
}
209-
}
210-
return NULL;
211-
}
212-
```
213-
214184
<!-- tabs:end -->
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216186
<!-- solution:end -->

‎solution/1400-1499/1436.Destination City/Solution.c

Lines changed: 0 additions & 15 deletions
This file was deleted.

‎solution/1400-1499/1436.Destination City/Solution.cpp

Lines changed: 5 additions & 5 deletions
Original file line numberDiff line numberDiff line change
@@ -5,11 +5,11 @@ class Solution {
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for (auto& p : paths) {
66
s.insert(p[0]);
77
}
8-
for (auto& p : paths) {
9-
if (!s.count(p[1])) {
10-
return p[1];
8+
for (int i = 0;; ++i) {
9+
auto b = paths[i][1];
10+
if (!s.contains(b)) {
11+
return b;
1112
}
1213
}
13-
return "";
1414
}
15-
};
15+
};

‎solution/1400-1499/1436.Destination City/Solution.java

Lines changed: 5 additions & 5 deletions
Original file line numberDiff line numberDiff line change
@@ -4,11 +4,11 @@ public String destCity(List<List<String>> paths) {
44
for (var p : paths) {
55
s.add(p.get(0));
66
}
7-
for (var p : paths) {
8-
if (!s.contains(p.get(1))) {
9-
return p.get(1);
7+
for (int i = 0;; ++i) {
8+
var b = paths.get(i).get(1);
9+
if (!s.contains(b)) {
10+
return b;
1011
}
1112
}
12-
return "";
1313
}
14-
}
14+
}

‎solution/1400-1499/1436.Destination City/Solution.js

Lines changed: 2 additions & 10 deletions
Original file line numberDiff line numberDiff line change
@@ -3,14 +3,6 @@
33
* @return {string}
44
*/
55
var destCity = function (paths) {
6-
const s = new Set();
7-
for (const [a, _] of paths) {
8-
s.add(a);
9-
}
10-
for (const [_, b] of paths) {
11-
if (!s.has(b)) {
12-
return b;
13-
}
14-
}
15-
return '';
6+
const s = new Set(paths.map(([a, _]) => a));
7+
return paths.find(([_, b]) => !s.has(b))[1];
168
};

‎solution/1400-1499/1436.Destination City/Solution.rs

Lines changed: 6 additions & 7 deletions
Original file line numberDiff line numberDiff line change
@@ -1,12 +1,11 @@
11
use std::collections::HashSet;
2+
23
impl Solution {
34
pub fn dest_city(paths: Vec<Vec<String>>) -> String {
4-
let set = paths.iter().map(|v| &v[0]).collect::<HashSet<&String>>();
5-
for path in paths.iter() {
6-
if !set.contains(&path[1]) {
7-
return path[1].clone();
8-
}
9-
}
10-
String::new()
5+
let s = paths
6+
.iter()
7+
.map(|p| p[0].clone())
8+
.collect::<HashSet<String>>();
9+
paths.into_iter().find(|p| !s.contains(&p[1])).unwrap()[1].clone()
1110
}
1211
}

‎solution/1400-1499/1436.Destination City/Solution.ts

Lines changed: 2 additions & 7 deletions
Original file line numberDiff line numberDiff line change
@@ -1,9 +1,4 @@
11
function destCity(paths: string[][]): string {
2-
const set = new Set(paths.map(([a]) => a));
3-
for (const [_, b] of paths) {
4-
if (!set.has(b)) {
5-
return b;
6-
}
7-
}
8-
return '';
2+
const s = new Set<string>(paths.map(([a, _]) => a));
3+
return paths.find(([_, b]) => !s.has(b))![1];
94
}

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