diff --git a/problems/0134.加油站.md b/problems/0134.加油站.md index dfed2d96..9b660ea0 100644 --- a/problems/0134.加油站.md +++ b/problems/0134.加油站.md @@ -240,6 +240,25 @@ class Solution: ``` Go: +```go +func canCompleteCircuit(gas []int, cost []int) int { + curSum := 0 + totalSum := 0 + start := 0 + for i := 0; i < len(gas); i++ { + curSum += gas[i] - cost[i] + totalSum += gas[i] - cost[i] + if curSum < 0 { + start = i+1 + curSum = 0 + } + } + if totalSum < 0 { + return -1 + } + return start +} +``` Javascript: ```Javascript diff --git a/problems/0209.长度最小的子数组.md b/problems/0209.长度最小的子数组.md index 90280451..42687514 100644 --- a/problems/0209.长度最小的子数组.md +++ b/problems/0209.长度最小的子数组.md @@ -109,7 +109,7 @@ public: }; ``` -时间复杂度:$O(n)$ +时间复杂度:$O(n)$ 空间复杂度:$O(1)$ **一些录友会疑惑为什么时间复杂度是O(n)**。 @@ -118,8 +118,8 @@ public: ## 相关题目推荐 -* 904.水果成篮 -* 76.最小覆盖子串 +* [904.水果成篮](https://leetcode-cn.com/problems/fruit-into-baskets/) +* [76.最小覆盖子串](https://leetcode-cn.com/problems/minimum-window-substring/) diff --git a/problems/0235.二叉搜索树的最近公共祖先.md b/problems/0235.二叉搜索树的最近公共祖先.md index 15ff7af4..d78db42a 100644 --- a/problems/0235.二叉搜索树的最近公共祖先.md +++ b/problems/0235.二叉搜索树的最近公共祖先.md @@ -265,6 +265,54 @@ class Solution: else: return root ``` Go: +> BSL法 + +```go +/** + * Definition for a binary tree node. + * type TreeNode struct { + * Val int + * Left *TreeNode + * Right *TreeNode + * } + */ +//利用BSL的性质(前序遍历有序) +func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode { + if root==nil{return nil} + if root.Val>p.Val&&root.Val>q.Val{//当前节点的值大于给定的值,则说明满足条件的在左边 + return lowestCommonAncestor(root.Left,p,q) + }else if root.Val 普通法 + +```go +/** + * Definition for a binary tree node. + * type TreeNode struct { + * Val int + * Left *TreeNode + * Right *TreeNode + * } + */ +//递归会将值层层返回 +func lowestCommonAncestor(root, p, q *TreeNode) *TreeNode { + //终止条件 + if root==nil||root.Val==p.Val||root.Val==q.Val{return root}//最后为空或者找到一个值时,就返回这个值 + //后序遍历 + findLeft:=lowestCommonAncestor(root.Left,p,q) + findRight:=lowestCommonAncestor(root.Right,p,q) + //处理单层逻辑 + if findLeft!=nil&&findRight!=nil{return root}//说明在root节点的两边 + if findLeft==nil{//左边没找到,就说明在右边找到了 + return findRight + }else {return findLeft} +} +``` + diff --git a/problems/0435.无重叠区间.md b/problems/0435.无重叠区间.md index 248526a1..23ee9f94 100644 --- a/problems/0435.无重叠区间.md +++ b/problems/0435.无重叠区间.md @@ -228,7 +228,27 @@ class Solution: Go: +Javascript: +```Javascript +var eraseOverlapIntervals = function(intervals) { + intervals.sort((a, b) => { + return a[1] - b[1] + }) + let count = 1 + let end = intervals[0][1] + + for(let i = 1; i < intervals.length; i++) { + let interval = intervals[i] + if(interval[0] >= right) { + end = interval[1] + count += 1 + } + } + + return intervals.length - count +}; +``` ----------------------- diff --git a/problems/0501.二叉搜索树中的众数.md b/problems/0501.二叉搜索树中的众数.md index dfd589ce..3ca7d892 100644 --- a/problems/0501.二叉搜索树中的众数.md +++ b/problems/0501.二叉搜索树中的众数.md @@ -428,7 +428,100 @@ class Solution: return self.res ``` Go: +暴力法(非BSL) +```go +/** + * Definition for a binary tree node. + * type TreeNode struct { + * Val int + * Left *TreeNode + * Right *TreeNode + * } + */ +func findMode(root *TreeNode) []int { + var history map[int]int + var maxValue int + var maxIndex int + var result []int + history=make(map[int]int) + traversal(root,history) + for k,value:=range history{ + if value>maxValue{ + maxValue=value + maxIndex=k + } + } + for k,value:=range history{ + if value==history[maxIndex]{ + result=append(result,k) + } + } + return result +} +func traversal(root *TreeNode,history map[int]int){ + if root.Left!=nil{ + traversal(root.Left,history) + } + if value,ok:=history[root.Val];ok{ + history[root.Val]=value+1 + }else{ + history[root.Val]=1 + } + if root.Right!=nil{ + traversal(root.Right,history) + } +} +``` + +计数法BSL(此代码在执行代码里能执行,但提交后报错,不知为何,思路是对的) + +```go +/** + * Definition for a binary tree node. + * type TreeNode struct { + * Val int + * Left *TreeNode + * Right *TreeNode + * } + */ + var count,maxCount int //统计计数 +func findMode(root *TreeNode) []int { + var result []int + var pre *TreeNode //前指针 + if root.Left==nil&&root.Right==nil{ + result=append(result,root.Val) + return result + } + traversal(root,&result,pre) + return result +} +func traversal(root *TreeNode,result *[]int,pre *TreeNode){//遍历统计 + //如果BSL中序遍历相邻的两个节点值相同,则统计频率;如果不相同,依据BSL中序遍历排好序的性质,重新计数 + if pre==nil{ + count=1 + }else if pre.Val==root.Val{ + count++ + }else { + count=1 + } + //如果统计的频率等于最大频率,则加入结果集;如果统计的频率大于最大频率,更新最大频率且重新将结果加入新的结果集中 + if count==maxCount{ + *result=append(*result,root.Val) + }else if count>maxCount{ + maxCount=count//重新赋值maxCount + *result=[]int{}//清空result中的内容 + *result=append(*result,root.Val) + } + pre=root//保存上一个的节点 + if root.Left!=nil{ + traversal(root.Left,result,pre) + } + if root.Right!=nil{ + traversal(root.Right,result,pre) + } +} +``` diff --git a/problems/0763.划分字母区间.md b/problems/0763.划分字母区间.md index bcdd71dc..1e2fcc03 100644 --- a/problems/0763.划分字母区间.md +++ b/problems/0763.划分字母区间.md @@ -128,7 +128,26 @@ class Solution: Go: - +Javascript: +```Javascript +var partitionLabels = function(s) { + let hash = {} + for(let i = 0; i < s.length; i++) { + hash[s[i]] = i + } + let result = [] + let left = 0 + let right = 0 + for(let i = 0; i < s.length; i++) { + right = Math.max(right, hash[s[i]]) + if(right === i) { + result.push(right - left + 1) + left = i + 1 + } + } + return result +}; +``` ----------------------- diff --git a/problems/1005.K次取反后最大化的数组和.md b/problems/1005.K次取反后最大化的数组和.md index 387de147..c3e99f7e 100644 --- a/problems/1005.K次取反后最大化的数组和.md +++ b/problems/1005.K次取反后最大化的数组和.md @@ -138,6 +138,30 @@ class Solution: ``` Go: +```Go +func largestSumAfterKNegations(nums []int, K int) int { + sort.Slice(nums, func(i, j int) bool { + return math.Abs(float64(nums[i])) > math.Abs(float64(nums[j])) + }) + + for i := 0; i < len(nums); i++ { + if K > 0 && nums[i] < 0 { + nums[i] = -nums[i] + K-- + } + } + + if K%2 == 1 { + nums[len(nums)-1] = -nums[len(nums)-1] + } + + result := 0 + for i := 0; i < len(nums); i++ { + result += nums[i] + } + return result +} +``` Javascript: diff --git a/problems/二叉树的递归遍历.md b/problems/二叉树的递归遍历.md index f2072e30..68f17257 100644 --- a/problems/二叉树的递归遍历.md +++ b/problems/二叉树的递归遍历.md @@ -221,12 +221,12 @@ class Solution: Go: 前序遍历: -``` +```go func PreorderTraversal(root *TreeNode) (res []int) { var traversal func(node *TreeNode) traversal = func(node *TreeNode) { if node == nil { - return + return } res = append(res,node.Val) traversal(node.Left) @@ -239,12 +239,12 @@ func PreorderTraversal(root *TreeNode) (res []int) { ``` 中序遍历: -``` +```go func InorderTraversal(root *TreeNode) (res []int) { var traversal func(node *TreeNode) traversal = func(node *TreeNode) { if node == nil { - return + return } traversal(node.Left) res = append(res,node.Val) @@ -256,12 +256,12 @@ func InorderTraversal(root *TreeNode) (res []int) { ``` 后序遍历: -``` +```go func PostorderTraversal(root *TreeNode) (res []int) { var traversal func(node *TreeNode) traversal = func(node *TreeNode) { if node == nil { - return + return } traversal(node.Left) traversal(node.Right) diff --git a/problems/背包理论基础01背包-2.md b/problems/背包理论基础01背包-2.md index e85d31b4..48275908 100644 --- a/problems/背包理论基础01背包-2.md +++ b/problems/背包理论基础01背包-2.md @@ -5,6 +5,7 @@

欢迎大家参与本项目,贡献其他语言版本的代码,拥抱开源,让更多学习算法的小伙伴们收益!

+ # 动态规划:关于01背包问题,你该了解这些!(滚动数组) 昨天[动态规划:关于01背包问题,你该了解这些!](https://mp.weixin.qq.com/s/FwIiPPmR18_AJO5eiidT6w)中是用二维dp数组来讲解01背包。 @@ -35,7 +36,7 @@ **其实可以发现如果把dp[i - 1]那一层拷贝到dp[i]上,表达式完全可以是:dp[i][j] = max(dp[i][j], dp[i][j - weight[i]] + value[i]);** -**于其把dp[i - 1]这一层拷贝到dp[i]上,不如只用一个一维数组了**,只用dp[j](一维数组,也可以理解是一个滚动数组)。 +**与其把dp[i - 1]这一层拷贝到dp[i]上,不如只用一个一维数组了**,只用dp[j](一维数组,也可以理解是一个滚动数组)。 这就是滚动数组的由来,需要满足的条件是上一层可以重复利用,直接拷贝到当前层。 @@ -214,7 +215,7 @@ int main() { Java: ```java - public static void main(String[] args) { + public static void main(String[] args) { int[] weight = {1, 3, 4}; int[] value = {15, 20, 30}; int bagWight = 4; @@ -242,7 +243,24 @@ Java: Python: +```python +def test_1_wei_bag_problem(): + weight = [1, 3, 4] + value = [15, 20, 30] + bag_weight = 4 + # 初始化: 全为0 + dp = [0] * (bag_weight + 1) + # 先遍历物品, 再遍历背包容量 + for i in range(len(weight)): + for j in range(bag_weight, weight[i] - 1, -1): + # 递归公式 + dp[j] = max(dp[j], dp[j - weight[i]] + value[i]) + + print(dp) + +test_1_wei_bag_problem() +``` Go: ```go