Merge branch 'master' of github.com:youngyangyang04/leetcode-master

This commit is contained in:
programmercarl
2023-09-19 16:21:48 +08:00
9 changed files with 317 additions and 9 deletions

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@ -694,6 +694,44 @@ object Solution {
}
```
### Ruby
```ruby
def letter_combinations(digits)
letter_map = {
2 => ['a','b','c'],
3 => ['d','e','f'],
4 => ['g','h','i'],
5 => ['j','k','l'],
6 => ['m','n','o'],
7 => ['p','q','r','s'],
8 => ['t','u','v'],
9 => ['w','x','y','z']
}
result = []
path = []
return result if digits.size == 0
backtracking(result, letter_map, digits.split(''), path, 0)
result
end
def backtracking(result, letter_map, digits, path, index)
if path.size == digits.size
result << path.join('')
return
end
hash[digits[index].to_i].each do |chr|
path << chr
#index + 1代表处理下一个数字
backtracking(result, letter_map, digits, path, index + 1)
#回溯,撤销处理过的数字
path.pop
end
end
```
<p align="center">
<a href="https://programmercarl.com/other/kstar.html" target="_blank">

View File

@ -764,6 +764,35 @@ object Solution {
}
```
### Ruby
```ruby
def combine(n, k)
result = []
path = []
backtracking(result, path, n, 1, k)
return result
end
#剪枝优化
def backtracking(result, path, n, j, k)
if path.size == k
result << path.map {|item| item}
return
end
for i in j..(n-(k - path.size)) + 1
#处理节点
path << i
backtracking(result, path, n, i + 1, k)
#回溯,撤销处理过的节点
path.pop
end
end
```
<p align="center">
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>

View File

@ -311,7 +311,7 @@ class Solution {
```
```java
//解法三:双反转+移位,在原始数组上进行反转。空间复杂度O(1)
//解法三:双反转+移位,String 的 toCharArray() 方法底层会 new 一个和原字符串相同大小的 char 数组,空间复杂度O(n)
class Solution {
/**
* 思路:

View File

@ -239,6 +239,42 @@ class Solution:
visited[next_i][next_j] = True
```
### JavaScript
```javascript
var numIslands = function (grid) {
let dir = [[0, 1], [1, 0], [-1, 0], [0, -1]]; // 四个方向
let bfs = (grid, visited, x, y) => {
let queue = [];
queue.push([x, y]);
visited[x][y] = true;
while (queue.length) {
let top = queue.shift();//取出队列头部元素
console.log(top)
for (let i = 0; i < 4; i++) {
let nextX = top[0] + dir[i][0]
let nextY = top[1] + dir[i][1]
if (nextX < 0 || nextX >= grid.length || nextY < 0 || nextY >= grid[0].length)
continue;
if (!visited[nextX][nextY] && grid[nextX][nextY] === "1") {
queue.push([nextX, nextY])
visited[nextX][nextY] = true
}
}
}
}
let visited = new Array(grid.length).fill().map(() => Array(grid[0].length).fill(false))
let res = 0
for (let i = 0; i < grid.length; i++) {
for (let j = 0; j < grid[i].length; j++) {
if (!visited[i][j] && grid[i][j] === "1") {
++res;
bfs(grid, visited, i, j);
}
}
}
return res
};
```
### Rust

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@ -219,7 +219,7 @@ class Solution {
}
```
Python:
### Python:
```python
# 版本一
@ -278,8 +278,91 @@ class Solution:
return result
```
### JavaScript
Rust:
```javascript
var numIslands = function (grid) {
let dir = [[0, 1], [1, 0], [-1, 0], [0, -1]]; // 四个方向
let dfs = (grid, visited, x, y) => {
for (let i = 0; i < 4; i++) {
let nextX = x + dir[i][0]
let nextY = y + dir[i][1]
if (nextX < 0 || nextX >= grid.length || nextY < 0 || nextY >= grid[0].length)
continue;
if (!visited[nextX][nextY] && grid[nextX][nextY] === "1") {
visited[nextX][nextY] = true
dfs(grid,visited,nextX,nextY)
}
}
}
let visited = new Array(grid.length).fill().map(() => Array(grid[0].length).fill(false))
let res = 0
for (let i = 0; i < grid.length; i++) {
for (let j = 0; j < grid[i].length; j++) {
if (!visited[i][j] && grid[i][j] === "1") {
++res;
visited[i][j] = true;
dfs(grid, visited, i, j);
}
}
}
return res
};
```
### Go
```go
func numIslands(grid [][]byte) int {
// 用1标记已访问
visited := make([][]int, len(grid))
for i := 0; i < len(visited); i++{
visited[i] = make([]int, len(grid[0]))
}
var bfs func(x, y int)
bfs = func(x, y int){
stack := make([][]int, 0)
stack = append(stack, []int{x, y})
moveX := []int{1, -1, 0, 0}
moveY := []int{0, 0, 1, -1}
for len(stack) != 0{
node := stack[len(stack) - 1]
stack = stack[:len(stack) - 1]
for i := 0; i < 4; i++{
dx := moveX[i] + node[0]
dy := moveY[i] + node[1]
if dx < 0 || dx >= len(grid) || dy < 0 || dy >= len(grid[0]) || visited[dx][dy] == 1{
continue
}
visited[dx][dy] = 1
if grid[dx][dy] == '1'{
stack = append(stack, []int{dx,dy})
}
}
}
}
result := 0
for i := 0; i < len(grid); i++{
for j := 0; j < len(grid[0]); j++{
if visited[i][j] == 0 && grid[i][j] == '1'{
bfs(i, j)
visited[i][j] = 1
result++
}
}
}
return result
}
```
### Rust:
```rust

View File

@ -454,13 +454,34 @@ class MyStack:
def top(self) -> int:
"""
写法一:
1. 首先确认不空
2. 我们仅有in会存放数据所以返回第一个即可
2. 我们仅有in会存放数据所以返回第一个即可(这里实际上用到了栈)
写法二:
1. 首先确认不空
2. 因为队列的特殊性FIFO所以我们只有在pop()的时候才会使用queue_out
3. 先把queue_in中的所有元素除了最后一个依次出列放进queue_out
4. 交换in和out此时out里只有一个元素
5. 把out中的pop出来即是原队列的最后一个并使用temp变量暂存
6. 把temp追加到queue_in的末尾
"""
# 写法一:
# if self.empty():
# return None
# return self.queue_in[-1] # 这里实际上用到了栈因为直接获取了queue_in的末尾元素
# 写法二:
if self.empty():
return None
for i in range(len(self.queue_in) - 1):
self.queue_out.append(self.queue_in.popleft())
return self.queue_in[-1]
self.queue_in, self.queue_out = self.queue_out, self.queue_in
temp = self.queue_out.popleft()
self.queue_in.append(temp)
return temp
def empty(self) -> bool:
@ -488,9 +509,19 @@ class MyStack:
return self.que.popleft()
def top(self) -> int:
# 写法一:
# if self.empty():
# return None
# return self.que[-1]
# 写法二:
if self.empty():
return None
return self.que[-1]
for i in range(len(self.que)-1):
self.que.append(self.que.popleft())
temp = self.que.popleft()
self.que.append(temp)
return temp
def empty(self) -> bool:
return not self.que

View File

@ -391,6 +391,44 @@ class Solution:
self.dfs(grid, visited, new_x, new_y)
```
### JavaScript
```javascript
var maxAreaOfIsland = function (grid) {
let dir = [[0, 1], [1, 0], [-1, 0], [0, -1]]; // 四个方向
let visited = new Array(grid.length).fill().map(() => Array(grid[0].length).fill(false))
let dfs = (grid, visited, x, y, m) => {
for (let i = 0; i < 4; i++) {
let nextX = x + dir[i][0]
let nextY = y + dir[i][1]
if (nextX < 0 || nextX >= grid.length || nextY < 0 || nextY >= grid[0].length)
continue;
if (!visited[nextX][nextY] && grid[nextX][nextY] === 1) {
visited[nextX][nextY] = true
m = dfs(grid, visited, nextX, nextY,m+1)
}
}
return m
}
let max = 0
for (let i = 0; i < grid.length; i++) {
for (let j = 0; j < grid[i].length; j++) {
if (!visited[i][j] && grid[i][j] === 1) {
// 深度优先
visited[i][j] = true;
let m = dfs(grid, visited, i, j, 1);
if (m > max) max = m;
}
}
}
return max
};
```
### Rust
dfs: 版本一
@ -530,7 +568,6 @@ impl Solution {
}
}
```
<p align="center">
<a href="https://programmercarl.com/other/kstar.html" target="_blank">
<img src="../pics/网站星球宣传海报.jpg" width="1000"/>

View File

@ -217,8 +217,61 @@ class Solution:
self.path.pop() # 回溯
```
### JavaScript
```javascript
var allPathsSourceTarget = function(graph) {
let res=[],path=[]
function dfs(graph,start){
if(start===graph.length-1){
res.push([...path])
return;
}
for(let i=0;i<graph[start].length;i++){
path.push(graph[start][i])
dfs(graph,graph[start][i])
path.pop()
}
}
path.push(0)
dfs(graph,0)
return res
};
```
### Go
```go
func allPathsSourceTarget(graph [][]int) [][]int {
result := make([][]int, 0)
var dfs func(path []int, step int)
dfs = func(path []int, step int){
// 从0遍历到length-1
if step == len(graph) - 1{
tmp := make([]int, len(path))
copy(tmp, path)
result = append(result, tmp)
return
}
for i := 0; i < len(graph[step]); i++{
next := append(path, graph[step][i])
dfs(next, graph[step][i])
}
}
// 从0开始开始push 0进去
dfs([]int{0}, 0)
return result
}
```
### Rust
```rust
impl Solution {
pub fn all_paths_source_target(graph: Vec<Vec<i32>>) -> Vec<Vec<i32>> {

View File

@ -120,8 +120,9 @@ class Solution {
```
```java
//解法二空间复杂度O(1)。用原始数组来进行反转操作
//思路为:先整个字符串反转,再反转前面的,最后反转后面 n 个
// 解法二
// 空间复杂度O(n)。String 的 toCharArray() 方法底层会 new 一个和原字符串相同大小的 char 数组
// 思路为:先整个字符串反转,再反转前面的,最后反转后面 n 个
class Solution {
public String reverseLeftWords(String s, int n) {
char[] chars = s.toCharArray();