diff --git a/problems/0001.两数之和.md b/problems/0001.两数之和.md index 41a95daf..02e9996f 100644 --- a/problems/0001.两数之和.md +++ b/problems/0001.两数之和.md @@ -135,6 +135,21 @@ func twoSum(nums []int, target int) []int { } ``` +```go +// 使用map方式解题，降低时间复杂度 +func twoSum(nums []int, target int) []int { + m := make(map[int]int) + for index, val := range nums { + if preIndex, ok := m[target-val]; ok { + return []int{preIndex, index} + } else { + m[val] = index + } + } + return []int{} +} +``` + Rust ```rust diff --git a/problems/0028.实现strStr.md b/problems/0028.实现strStr.md index 6f257d5e..04508440 100644 --- a/problems/0028.实现strStr.md +++ b/problems/0028.实现strStr.md @@ -721,10 +721,98 @@ class Solution: Go： +```go +// 方法一:前缀表使用减1实现 + +// getNext 构造前缀表next +// params: +// next 前缀表数组 +// s 模式串 +func getNext(next []int, s string) { + j := -1 // j表示 最长相等前后缀长度 + next[0] = j + + for i := 1; i < len(s); i++ { + for j >= 0 && s[i] != s[j+1] { + j = next[j] // 回退前一位 + } + if s[i] == s[j+1] { + j++ + } + next[i] = j // next[i]是i（包括i）之前的最长相等前后缀长度 + } +} +func strStr(haystack string, needle string) int { + if len(needle) == 0 { + return 0 + } + next := make([]int, len(needle)) + getNext(next, needle) + j := -1 // 模式串的起始位置 next为-1 因此也为-1 + for i := 0; i < len(haystack); i++ { + for j >= 0 && haystack[i] != needle[j+1] { + j = next[j] // 寻找下一个匹配点 + } + if haystack[i] == needle[j+1] { + j++ + } + if j == len(needle)-1 { // j指向了模式串的末尾 + return i - len(needle) + 1 + } + } + return -1 +} +``` + +```go +// 方法二: 前缀表无减一或者右移 + +// getNext 构造前缀表next +// params: +// next 前缀表数组 +// s 模式串 +func getNext(next []int, s string) { + j := 0 + next[0] = j + for i := 1; i < len(s); i++ { + for j > 0 && s[i] != s[j] { + j = next[j-1] + } + if s[i] == s[j] { + j++ + } + next[i] = j + } +} +func strStr(haystack string, needle string) int { + n := len(needle) + if n == 0 { + return 0 + } + j := 0 + next := make([]int, n) + getNext(next, needle) + for i := 0; i < len(haystack); i++ { + for j > 0 && haystack[i] != needle[j] { + j = next[j-1] // 回退到j的前一位 + } + if haystack[i] == needle[j] { + j++ + } + if j == n { + return i - n + 1 + } + } + return -1 +} +``` + + + ----------------------- * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw) * B站视频：[代码随想录](https://space.bilibili.com/525438321) * 知识星球：[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ) -
+
\ No newline at end of file diff --git a/problems/0037.解数独.md b/problems/0037.解数独.md index d5df95e8..4eb60704 100644 --- a/problems/0037.解数独.md +++ b/problems/0037.解数独.md @@ -287,7 +287,39 @@ class Solution { ``` Python： - +```python3 +class Solution: + def solveSudoku(self, board: List[List[str]]) -> None: + """ + Do not return anything, modify board in-place instead. + """ + def backtrack(board): + for i in range(len(board)): #遍历行 + for j in range(len(board[0])): #遍历列 + if board[i][j] != ".": continue + for k in range(1,10): #(i, j) 这个位置放k是否合适 + if isValid(i,j,k,board): + board[i][j] = str(k) #放置k + if backtrack(board): return True #如果找到合适一组立刻返回 + board[i][j] = "." #回溯，撤销k + return False #9个数都试完了，都不行，那么就返回false + return True #遍历完没有返回false，说明找到了合适棋盘位置了 + def isValid(row,col,val,board): + for i in range(9): #判断行里是否重复 + if board[row][i] == str(val): + return False + for j in range(9): #判断列里是否重复 + if board[j][col] == str(val): + return False + startRow = (row // 3) * 3 + startcol = (col // 3) * 3 + for i in range(startRow,startRow + 3): #判断9方格里是否重复 + for j in range(startcol,startcol + 3): + if board[i][j] == str(val): + return False + return True + backtrack(board) +``` Go： diff --git a/problems/0056.合并区间.md b/problems/0056.合并区间.md index eb8e7bff..d4ffc554 100644 --- a/problems/0056.合并区间.md +++ b/problems/0056.合并区间.md @@ -141,16 +141,7 @@ Java： class Solution { public int[][] merge(int[][] intervals) { List res = new LinkedList<>(); - Arrays.sort(intervals, new Comparator() { - @Override - public int compare(int[] o1, int[] o2) { - if (o1[0] != o2[0]) { - return Integer.compare(o1[0],o2[0]); - } else { - return Integer.compare(o1[1],o2[1]); - } - } - }); + Arrays.sort(intervals, (o1, o2) -> Integer.compare(o1[0], o2[0])); int start = intervals[0][0]; for (int i = 1; i < intervals.length; i++) { diff --git a/problems/0101.对称二叉树.md b/problems/0101.对称二叉树.md index d8797d30..b58cef2e 100644 --- a/problems/0101.对称二叉树.md +++ b/problems/0101.对称二叉树.md @@ -363,6 +363,54 @@ Python： Go： +```go +/** + * Definition for a binary tree node. + * type TreeNode struct { + * Val int + * Left *TreeNode + * Right *TreeNode + * } + */ +// 递归 +func defs(left *TreeNode, right *TreeNode) bool { + if left == nil && right == nil { + return true; + }; + if left == nil || right == nil { + return false; + }; + if left.Val != right.Val { + return false; + } + return defs(left.Left, right.Right) && defs(right.Left, left.Right); +} +func isSymmetric(root *TreeNode) bool { + return defs(root.Left, root.Right); +} + +// 迭代 +func isSymmetric(root *TreeNode) bool { + var queue []*TreeNode; + if root != nil { + queue = append(queue, root.Left, root.Right); + } + for len(queue) > 0 { + left := queue[0]; + right := queue[1]; + queue = queue[2:]; + if left == nil && right == nil { + continue; + } + if left == nil || right == nil || left.Val != right.Val { + return false; + }; + queue = append(queue, left.Left, right.Right, right.Left, left.Right); + } + return true; +} +``` + JavaScript ```javascript diff --git a/problems/0104.二叉树的最大深度.md b/problems/0104.二叉树的最大深度.md index 756afb68..5f0fe411 100644 --- a/problems/0104.二叉树的最大深度.md +++ b/problems/0104.二叉树的最大深度.md @@ -284,6 +284,55 @@ Python： Go： +```go +/** + * Definition for a binary tree node. + * type TreeNode struct { + * Val int + * Left *TreeNode + * Right *TreeNode + * } + */ +func max (a, b int) int { + if a > b { + return a; + } + return b; +} +// 递归 +func maxDepth(root *TreeNode) int { + if root == nil { + return 0; + } + return max(maxDepth(root.Left), maxDepth(root.Right)) + 1; +} + +// 遍历 +func maxDepth(root *TreeNode) int { + levl := 0; + queue := make([]*TreeNode, 0); + if root != nil { + queue = append(queue, root); + } + for l := len(queue); l > 0; { + for ;l > 0;l-- { + node := queue[0]; + if node.Left != nil { + queue = append(queue, node.Left); + } + if node.Right != nil { + queue = append(queue, node.Right); + } + queue = queue[1:]; + } + levl++; + l = len(queue); + } + return levl; +} + +``` + JavaScript ```javascript diff --git a/problems/0111.二叉树的最小深度.md b/problems/0111.二叉树的最小深度.md index 8ee15eac..48795722 100644 --- a/problems/0111.二叉树的最小深度.md +++ b/problems/0111.二叉树的最小深度.md @@ -301,6 +301,64 @@ class Solution: Go： +```go +/** + * Definition for a binary tree node. + * type TreeNode struct { + * Val int + * Left *TreeNode + * Right *TreeNode + * } + */ +func min(a, b int) int { + if a < b { + return a; + } + return b; +} +// 递归 +func minDepth(root *TreeNode) int { + if root == nil { + return 0; + } + if root.Left == nil && root.Right != nil { + return 1 + minDepth(root.Right); + } + if root.Right == nil && root.Left != nil { + return 1 + minDepth(root.Left); + } + return min(minDepth(root.Left), minDepth(root.Right)) + 1; +} + +// 迭代 + +func minDepth(root *TreeNode) int { + dep := 0; + queue := make([]*TreeNode, 0); + if root != nil { + queue = append(queue, root); + } + for l := len(queue); l > 0; { + dep++; + for ; l > 0; l-- { + node := queue[0]; + if node.Left == nil && node.Right == nil { + return dep; + } + if node.Left != nil { + queue = append(queue, node.Left); + } + if node.Right != nil { + queue = append(queue, node.Right); + } + queue = queue[1:]; + } + l = len(queue); + } + return dep; +} +``` + JavaScript: diff --git a/problems/0112.路径总和.md b/problems/0112.路径总和.md index 4ccd8912..d810a046 100644 --- a/problems/0112.路径总和.md +++ b/problems/0112.路径总和.md @@ -345,6 +345,36 @@ class Solution { return hasPathSum(root.left, targetSum - root.val) || hasPathSum(root.right, targetSum - root.val); } } +``` +迭代 +```java +class Solution { + public boolean hasPathSum(TreeNode root, int targetSum) { + if(root==null)return false; + Stack stack1 = new Stack<>(); + Stack stack2 = new Stack<>(); + stack1.push(root);stack2.push(root.val); + while(!stack1.isEmpty()){ + int size = stack1.size(); + for(int i=0;i ratings[i - 1]) { + candys[i] = candys[i - 1] + 1 + } + } + + for(let i = ratings.length - 2; i >= 0; i--) { + if(ratings[i] > ratings[i + 1]) { + candys[i] = Math.max(candys[i], candys[i + 1] + 1) + } + } + + let count = candys.reduce((a, b) => { + return a + b + }) + + return count +}; +``` ----------------------- 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 512360fe..63499b71 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 @@ -318,9 +318,61 @@ class Solution { Python： - Go： +```go +import ( + "fmt" +) + +func reverseWords(s string) string { + //1.使用双指针删除冗余的空格 + slowIndex, fastIndex := 0, 0 + b := []byte(s) + //删除头部冗余空格 + for len(b) > 0 && fastIndex < len(b) && b[fastIndex] == ' ' { + fastIndex++ + } + //删除单词间冗余空格 + for ; fastIndex < len(b); fastIndex++ { + if fastIndex-1 > 0 && b[fastIndex-1] == b[fastIndex] && b[fastIndex] == ' ' { + continue + } + b[slowIndex] = b[fastIndex] + slowIndex++ + } + //删除尾部冗余空格 + if slowIndex-1 > 0 && b[slowIndex-1] == ' ' { + b = b[:slowIndex-1] + } else { + b = b[:slowIndex] + } + //2.反转整个字符串 + reverse(&b, 0, len(b)-1) + //3.反转单个单词 i单词开始位置，j单词结束位置 + i := 0 + for i < len(b) { + j := i + for ; j < len(b) && b[j] != ' '; j++ { + } + reverse(&b, i, j-1) + i = j + i++ + } + return string(b) +} + +func reverse(b *[]byte, left, right int) { + for left < right { + (*b)[left], (*b)[right] = (*b)[right], (*b)[left] + left++ + right-- + } +} +``` + + + @@ -328,4 +380,4 @@ Go： * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw) * B站视频：[代码随想录](https://space.bilibili.com/525438321) * 知识星球：[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ) -
+
\ No newline at end of file diff --git a/problems/0239.滑动窗口最大值.md b/problems/0239.滑动窗口最大值.md index ed17157a..47383a47 100644 --- a/problems/0239.滑动窗口最大值.md +++ b/problems/0239.滑动窗口最大值.md @@ -263,6 +263,41 @@ class Solution { ``` Python： +```python +class MyQueue: #单调队列（从大到小 + def __init__(self): + self.queue = [] #使用list来实现单调队列 + + #每次弹出的时候，比较当前要弹出的数值是否等于队列出口元素的数值，如果相等则弹出。 + #同时pop之前判断队列当前是否为空。 + def pop(self, value): + if self.queue and value == self.queue[0]: + self.queue.pop(0)#list.pop()时间复杂度为O(n),这里可以使用collections.deque() + + #如果push的数值大于入口元素的数值，那么就将队列后端的数值弹出，直到push的数值小于等于队列入口元素的数值为止。 + #这样就保持了队列里的数值是单调从大到小的了。 + def push(self, value): + while self.queue and value > self.queue[-1]: + self.queue.pop() + self.queue.append(value) + + #查询当前队列里的最大值 直接返回队列前端也就是front就可以了。 + def front(self): + return self.queue[0] + +class Solution: + def maxSlidingWindow(self, nums: List[int], k: int) -> List[int]: + que = MyQueue() + result = [] + for i in range(k): #先将前k的元素放进队列 + que.push(nums[i]) + result.append(que.front()) #result 记录前k的元素的最大值 + for i in range(k, len(nums)): + que.pop(nums[i - k]) #滑动窗口移除最前面元素 + que.push(nums[i]) #滑动窗口前加入最后面的元素 + result.append(que.front()) #记录对应的最大值 + return result +``` Go： diff --git a/problems/0347.前K个高频元素.md b/problems/0347.前K个高频元素.md index 841584b4..71af618e 100644 --- a/problems/0347.前K个高频元素.md +++ b/problems/0347.前K个高频元素.md @@ -162,7 +162,33 @@ class Solution { Python： - +```python +#时间复杂度：O(nlogk) +#空间复杂度：O(n) +import heapq +class Solution: + def topKFrequent(self, nums: List[int], k: int) -> List[int]: + #要统计元素出现频率 + map_ = {} #nums[i]:对应出现的次数 + for i in range(len(nums)): + map_[nums[i]] = map_.get(nums[i], 0) + 1 + + #对频率排序 + #定义一个小顶堆，大小为k + pri_que = [] #小顶堆 + + #用固定大小为k的小顶堆，扫面所有频率的数值 + for key, freq in map_.items(): + heapq.heappush(pri_que, (freq, key)) + if len(pri_que) > k: #如果堆的大小大于了K，则队列弹出，保证堆的大小一直为k + heapq.heappop(pri_que) + + #找出前K个高频元素，因为小顶堆先弹出的是最小的，所以倒叙来输出到数组 + result = [0] * k + for i in range(k-1, -1, -1): + result[i] = heapq.heappop(pri_que)[1] + return result +``` Go： diff --git a/problems/0376.摆动序列.md b/problems/0376.摆动序列.md index 6fa30cde..bbca5ea0 100644 --- a/problems/0376.摆动序列.md +++ b/problems/0376.摆动序列.md @@ -138,7 +138,21 @@ class Solution { ``` Python： - +```python +class Solution: + def wiggleMaxLength(self, nums: List[int]) -> int: + #贪心 求波峰数量 + 波谷数量 + if len(nums)<=1: + return len(nums) + cur, pre = 0,0 #当前一对差值，前一对差值 + count = 1#默认最右边有一个峰值 + for i in range(len(nums)-1): + cur = nums[i+1] - nums[i] + if((cur>0 and pre<=0) or (cur<0 and pre>=0)): + count += 1 + pre = cur + return count +``` Go： diff --git a/problems/0383.赎金信.md b/problems/0383.赎金信.md index fd8175db..23e2c5fd 100644 --- a/problems/0383.赎金信.md +++ b/problems/0383.赎金信.md @@ -166,6 +166,21 @@ class Solution(object): ``` Go： +```go +func canConstruct(ransomNote string, magazine string) bool { + record := make([]int, 26) + for _, v := range magazine { + record[v-'a']++ + } + for _, v := range ransomNote { + record[v-'a']-- + if record[v-'a'] < 0 { + return false + } + } + return true +} +``` javaScript: diff --git a/problems/0454.四数相加II.md b/problems/0454.四数相加II.md index 939ed20d..835178dd 100644 --- a/problems/0454.四数相加II.md +++ b/problems/0454.四数相加II.md @@ -153,6 +153,23 @@ class Solution(object): Go： +```go +func fourSumCount(nums1 []int, nums2 []int, nums3 []int, nums4 []int) int { + m := make(map[int]int) + count := 0 + for _, v1 := range nums1 { + for _, v2 := range nums2 { + m[v1+v2]++ + } + } + for _, v3 := range nums3 { + for _, v4 := range nums4 { + count += m[-v3-v4] + } + } + return count +} +``` javaScript: diff --git a/problems/0455.分发饼干.md b/problems/0455.分发饼干.md index b3714c43..ecf7f132 100644 --- a/problems/0455.分发饼干.md +++ b/problems/0455.分发饼干.md @@ -134,7 +134,21 @@ class Solution { ``` Python： - +```python +class Solution: + def findContentChildren(self, g: List[int], s: List[int]) -> int: + #先考虑胃口小的孩子 + g.sort() + s.sort() + i=j=0 + count = 0 + while(i bool: + if len(s) == 0: + return False + nxt = [0] * len(s) + self.getNext(nxt, s) + if nxt[-1] != -1 and len(s) % (len(s) - (nxt[-1] + 1)) == 0: + return True + return False + + def getNext(self, nxt, s): + nxt[0] = -1 + j = -1 + for i in range(1, len(s)): + while j >= 0 and s[i] != s[j+1]: + j = nxt[j] + if s[i] == s[j+1]: + j += 1 + nxt[i] = j + return nxt +``` + +前缀表（不减一）的代码实现 + +```python +class Solution: + def repeatedSubstringPattern(self, s: str) -> bool: + if len(s) == 0: + return False + nxt = [0] * len(s) + self.getNext(nxt, s) + if nxt[-1] != 0 and len(s) % (len(s) - nxt[-1]) == 0: + return True + return False + + def getNext(self, nxt, s): + nxt[0] = 0 + j = 0 + for i in range(1, len(s)): + while j > 0 and s[i] != s[j]: + j = nxt[j - 1] + if s[i] == s[j]: + j += 1 + nxt[i] = j + return nxt +``` Go： +这里使用了前缀表统一减一的实现方式 + +```go +func repeatedSubstringPattern(s string) bool { + n := len(s) + if n == 0 { + return false + } + next := make([]int, n) + j := -1 + next[0] = j + for i := 1; i < n; i++ { + for j >= 0 && s[i] != s[j+1] { + j = next[j] + } + if s[i] == s[j+1] { + j++ + } + next[i] = j + } + // next[n-1]+1 最长相同前后缀的长度 + if next[n-1] != -1 && n%(n-(next[n-1]+1)) == 0 { + return true + } + return false +} +``` + +前缀表（不减一）的代码实现 + +```go +func repeatedSubstringPattern(s string) bool { + n := len(s) + if n == 0 { + return false + } + j := 0 + next := make([]int, n) + next[0] = j + for i := 1; i < n; i++ { + for j > 0 && s[i] != s[j] { + j = next[j-1] + } + if s[i] == s[j] { + j++ + } + next[i] = j + } + // next[n-1] 最长相同前后缀的长度 + if next[n-1] != 0 && n%(n-next[n-1]) == 0 { + return true + } + return false +} +``` + + + @@ -194,4 +301,4 @@ Go： * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw) * B站视频：[代码随想录](https://space.bilibili.com/525438321) * 知识星球：[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ) -
+
\ No newline at end of file diff --git a/problems/0494.目标和.md b/problems/0494.目标和.md index c772a09d..c917ed5c 100644 --- a/problems/0494.目标和.md +++ b/problems/0494.目标和.md @@ -276,7 +276,35 @@ class Solution: ``` Go： - +```go +func findTargetSumWays(nums []int, target int) int { + sum := 0 + for _, v := range nums { + sum += v + } + if target > sum { + return 0 + } + if (sum+target)%2 == 1 { + return 0 + } + // 计算背包大小 + bag := (sum + target) / 2 + // 定义dp数组 + dp := make([]int, bag+1) + // 初始化 + dp[0] = 1 + // 遍历顺序 + for i := 0; i < len(nums); i++ { + for j := bag; j >= nums[i]; j-- { + //推导公式 + dp[j] += dp[j-nums[i]] + //fmt.Println(dp) + } + } + return dp[bag] +} +``` diff --git a/problems/1049.最后一块石头的重量II.md b/problems/1049.最后一块石头的重量II.md index f74600b1..c09e476a 100644 --- a/problems/1049.最后一块石头的重量II.md +++ b/problems/1049.最后一块石头的重量II.md @@ -191,7 +191,33 @@ class Solution: ``` Go： +```go +func lastStoneWeightII(stones []int) int { + // 15001 = 30 * 1000 /2 +1 + dp := make([]int, 15001) + // 求target + sum := 0 + for _, v := range stones { + sum += v + } + target := sum / 2 + // 遍历顺序 + for i := 0; i < len(stones); i++ { + for j := target; j >= stones[i]; j-- { + // 推导公式 + dp[j] = max(dp[j], dp[j-stones[i]]+stones[i]) + } + } + return sum - 2 * dp[target] +} +func max(a, b int) int { + if a > b { + return a + } + return b +} +``` diff --git a/problems/二叉树的统一迭代法.md b/problems/二叉树的统一迭代法.md index f4091ad5..533bdfe7 100644 --- a/problems/二叉树的统一迭代法.md +++ b/problems/二叉树的统一迭代法.md @@ -239,7 +239,78 @@ Java： ``` Python： +> 迭代法前序遍历 +```python +class Solution: + def preorderTraversal(self, root: TreeNode) -> List[int]: + result = [] + st= [] + if root: + st.append(root) + while st: + node = st.pop() + if node != None: + if node.right: #右 + st.append(node.right) + if node.left: #左 + st.append(node.left) + st.append(node) #中 + st.append(None) + else: + node = st.pop() + result.append(node.val) + return result +``` + +> 迭代法中序遍历 +```python +class Solution: + def inorderTraversal(self, root: TreeNode) -> List[int]: + result = [] + st = [] + if root: + st.append(root) + while st: + node = st.pop() + if node != None: + if node.right: #添加右节点（空节点不入栈） + st.append(node.right) + + st.append(node) #添加中节点 + st.append(None) #中节点访问过，但是还没有处理，加入空节点做为标记。 + + if node.left: #添加左节点（空节点不入栈） + st.append(node.left) + else: #只有遇到空节点的时候，才将下一个节点放进结果集 + node = st.pop() #重新取出栈中元素 + result.append(node.val) #加入到结果集 + return result +``` + +> 迭代法后序遍历 +```python +class Solution: + def postorderTraversal(self, root: TreeNode) -> List[int]: + result = [] + st = [] + if root: + st.append(root) + while st: + node = st.pop() + if node != None: + st.append(node) #中 + st.append(None) + + if node.right: #右 + st.append(node.right) + if node.left: #左 + st.append(node.left) + else: + node = st.pop() + result.append(node.val) + return result +``` Go： > 前序遍历统一迭代法 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 648546c1..39c8382c 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 @@ -23,7 +23,7 @@ https://leetcode-cn.com/problems/zuo-xuan-zhuan-zi-fu-chuan-lcof/ 示例 2： 输入: s = "lrloseumgh", k = 6 输出: "umghlrlose" -  + 限制： 1 <= k < s.length <= 10000 @@ -119,9 +119,31 @@ class Solution { ``` Python： - Go： +```go +func reverseLeftWords(s string, n int) string { + b := []byte(s) + // 1. 反转前n个字符 + // 2. 反转第n到end字符 + // 3. 反转整个字符 + reverse(b, 0, n-1) + reverse(b, n, len(b)-1) + reverse(b, 0, len(b)-1) + return string(b) +} +// 切片是引用传递 +func reverse(b []byte, left, right int){ + for left < right{ + b[left], b[right] = b[right],b[left] + left++ + right-- + } +} +``` + + + @@ -129,4 +151,4 @@ Go： * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw) * B站视频：[代码随想录](https://space.bilibili.com/525438321) * 知识星球：[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ) -
+
\ No newline at end of file diff --git a/problems/链表理论基础.md b/problems/链表理论基础.md index ea3b9098..252247c7 100644 --- a/problems/链表理论基础.md +++ b/problems/链表理论基础.md @@ -83,7 +83,7 @@ struct ListNode { 有同学说了，我不定义构造函数行不行，答案是可以的，C++默认生成一个构造函数。 -但是这个构造函数不会初始化任何成员变化，下面我来举两个例子： +但是这个构造函数不会初始化任何成员变量，下面我来举两个例子： 通过自己定义构造函数初始化节点：