diff --git a/problems/0017.电话号码的字母组合.md b/problems/0017.电话号码的字母组合.md index b3ba1e5e..90296efd 100644 --- a/problems/0017.电话号码的字母组合.md +++ b/problems/0017.电话号码的字母组合.md @@ -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 +```

diff --git a/problems/0077.组合.md b/problems/0077.组合.md index 8d448739..a7f00ffe 100644 --- a/problems/0077.组合.md +++ b/problems/0077.组合.md @@ -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 + +``` +

diff --git a/problems/0151.翻转字符串里的单词.md b/problems/0151.翻转字符串里的单词.md index 0c1a526f..111c07e4 100644 --- a/problems/0151.翻转字符串里的单词.md +++ b/problems/0151.翻转字符串里的单词.md @@ -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： diff --git a/problems/0200.岛屿数量.广搜版.md b/problems/0200.岛屿数量.广搜版.md index 8bbedb59..e8ed60db 100644 --- a/problems/0200.岛屿数量.广搜版.md +++ b/problems/0200.岛屿数量.广搜版.md @@ -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 diff --git a/problems/0200.岛屿数量.深搜版.md b/problems/0200.岛屿数量.深搜版.md index 905c0979..c7649971 100644 --- a/problems/0200.岛屿数量.深搜版.md +++ b/problems/0200.岛屿数量.深搜版.md @@ -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 diff --git a/problems/0225.用队列实现栈.md b/problems/0225.用队列实现栈.md index 3de300c7..d89bf44b 100644 --- a/problems/0225.用队列实现栈.md +++ b/problems/0225.用队列实现栈.md @@ -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 diff --git a/problems/0695.岛屿的最大面积.md b/problems/0695.岛屿的最大面积.md index 3dd181a3..74470ae5 100644 --- a/problems/0695.岛屿的最大面积.md +++ b/problems/0695.岛屿的最大面积.md @@ -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 { } } ``` -

diff --git a/problems/0797.所有可能的路径.md b/problems/0797.所有可能的路径.md index 9d14bd7c..bcce9314 100644 --- a/problems/0797.所有可能的路径.md +++ b/problems/0797.所有可能的路径.md @@ -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>) -> Vec> { diff --git a/problems/剑指Offer58-II.左旋转字符串.md b/problems/剑指Offer58-II.左旋转字符串.md index 82a968d4..43a0fe08 100644 --- a/problems/剑指Offer58-II.左旋转字符串.md +++ b/problems/剑指Offer58-II.左旋转字符串.md @@ -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();