diff --git a/problems/0112.路径总和.md b/problems/0112.路径总和.md index d810a046..54f79d1d 100644 --- a/problems/0112.路径总和.md +++ b/problems/0112.路径总和.md @@ -486,6 +486,92 @@ class Solution: Go： +> 112. 路径总和 + +```go +//递归法 +/** + * Definition for a binary tree node. + * type TreeNode struct { + * Val int + * Left *TreeNode + * Right *TreeNode + * } + */ +func hasPathSum(root *TreeNode, targetSum int) bool { + var flage bool //找没找到的标志 + if root==nil{ + return flage + } + pathSum(root,0,targetSum,&flage) + return flage +} +func pathSum(root *TreeNode, sum int,targetSum int,flage *bool){ + sum+=root.Val + if root.Left==nil&&root.Right==nil&&sum==targetSum{ + *flage=true + return + } + if root.Left!=nil&&!(*flage){//左节点不为空且还没找到 + pathSum(root.Left,sum,targetSum,flage) + } + if root.Right!=nil&&!(*flage){//右节点不为空且没找到 + pathSum(root.Right,sum,targetSum,flage) + } +} +``` + + + +> 113 递归法 + +```go +/** + * Definition for a binary tree node. + * type TreeNode struct { + * Val int + * Left *TreeNode + * Right *TreeNode + * } + */ +func pathSum(root *TreeNode, targetSum int) [][]int { + var result [][]int//最终结果 + if root==nil{ + return result + } + var sumNodes []int//经过路径的节点集合 + hasPathSum(root,&sumNodes,targetSum,&result) + return result +} +func hasPathSum(root *TreeNode,sumNodes *[]int,targetSum int,result *[][]int){ + *sumNodes=append(*sumNodes,root.Val) + if root.Left==nil&&root.Right==nil{//叶子节点 + fmt.Println(*sumNodes) + var sum int + var number int + for k,v:=range *sumNodes{//求该路径节点的和 + sum+=v + number=k + } + tempNodes:=make([]int,number+1)//新的nodes接受指针里的值，防止最终指针里的值发生变动，导致最后的结果都是最后一个sumNodes的值 + for k,v:=range *sumNodes{ + tempNodes[k]=v + } + if sum==targetSum{ + *result=append(*result,tempNodes) + } + } + if root.Left!=nil{ + hasPathSum(root.Left,sumNodes,targetSum,result) + *sumNodes=(*sumNodes)[:len(*sumNodes)-1]//回溯 + } + if root.Right!=nil{ + hasPathSum(root.Right,sumNodes,targetSum,result) + *sumNodes=(*sumNodes)[:len(*sumNodes)-1]//回溯 + } +} +``` + JavaScript： 0112.路径总和 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 47383a47..d108335b 100644 --- a/problems/0239.滑动窗口最大值.md +++ b/problems/0239.滑动窗口最大值.md @@ -332,6 +332,66 @@ func maxSlidingWindow(nums []int, k int) []int { ``` +```go +// 封装单调队列的方式解题 +type MyQueue struct { + queue []int +} + +func NewMyQueue() *MyQueue { + return &MyQueue{ + queue: make([]int, 0), + } +} + +func (m *MyQueue) Front() int { + return m.queue[0] +} + +func (m *MyQueue) Back() int { + return m.queue[len(m.queue)-1] +} + +func (m *MyQueue) Empty() bool { + return len(m.queue) == 0 +} + +func (m *MyQueue) Push(val int) { + for !m.Empty() && val > m.Back() { + m.queue = m.queue[:len(m.queue)-1] + } + m.queue = append(m.queue, val) +} + +func (m *MyQueue) Pop(val int) { + if !m.Empty() && val == m.Front() { + m.queue = m.queue[1:] + } +} + +func maxSlidingWindow(nums []int, k int) []int { + queue := NewMyQueue() + length := len(nums) + res := make([]int, 0) + // 先将前k个元素放入队列 + for i := 0; i < k; i++ { + queue.Push(nums[i]) + } + // 记录前k个元素的最大值 + res = append(res, queue.Front()) + + for i := k; i < length; i++ { + // 滑动窗口移除最前面的元素 + queue.Pop(nums[i-k]) + // 滑动窗口添加最后面的元素 + queue.Push(nums[i]) + // 记录最大值 + res = append(res, queue.Front()) + } + return res +} +``` + Javascript: ```javascript var maxSlidingWindow = function (nums, k) { diff --git a/problems/0242.有效的字母异位词.md b/problems/0242.有效的字母异位词.md index ba942f70..93bba44c 100644 --- a/problems/0242.有效的字母异位词.md +++ b/problems/0242.有效的字母异位词.md @@ -130,7 +130,27 @@ class Solution: return True ``` +Python写法二（没有使用数组作为哈希表，只是介绍defaultdict这样一种解题思路）： + +```python +class Solution: + def isAnagram(self, s: str, t: str) -> bool: + from collections import defaultdict + + s_dict = defaultdict(int) + t_dict = defaultdict(int) + + for x in s: + s_dict[x] += 1 + + for x in t: + t_dict[x] += 1 + + return s_dict == t_dict +``` + Go： + ```go func isAnagram(s string, t string) bool { if len(s)!=len(t){ diff --git a/problems/0343.整数拆分.md b/problems/0343.整数拆分.md index fefaa293..cf60575f 100644 --- a/problems/0343.整数拆分.md +++ b/problems/0343.整数拆分.md @@ -218,7 +218,7 @@ class Solution: # 假设对正整数 i 拆分出的第一个正整数是 j（1 <= j < i），则有以下两种方案： # 1) 将 i 拆分成 j 和 i−j 的和，且 i−j 不再拆分成多个正整数，此时的乘积是 j * (i-j) # 2) 将 i 拆分成 j 和 i−j 的和，且 i−j 继续拆分成多个正整数，此时的乘积是 j * dp[i-j] - for j in range(1, i): + for j in range(1, i - 1): dp[i] = max(dp[i], max(j * (i - j), j * dp[i - j])) return dp[n] ``` diff --git a/problems/0404.左叶子之和.md b/problems/0404.左叶子之和.md index e5daa9db..aa758367 100644 --- a/problems/0404.左叶子之和.md +++ b/problems/0404.左叶子之和.md @@ -226,6 +226,71 @@ class Solution: ``` Go： +> 递归法 + +```go +/** + * Definition for a binary tree node. + * type TreeNode struct { + * Val int + * Left *TreeNode + * Right *TreeNode + * } + */ +func sumOfLeftLeaves(root *TreeNode) int { + var res int + findLeft(root,&res) + return res +} +func findLeft(root *TreeNode,res *int){ + //左节点 + if root.Left!=nil&&root.Left.Left==nil&&root.Left.Right==nil{ + *res=*res+root.Left.Val + } + if root.Left!=nil{ + findLeft(root.Left,res) + } + if root.Right!=nil{ + findLeft(root.Right,res) + } +} +``` + +> 迭代法 + +```go +/** + * Definition for a binary tree node. + * type TreeNode struct { + * Val int + * Left *TreeNode + * Right *TreeNode + * } + */ +func sumOfLeftLeaves(root *TreeNode) int { + var res int + queue:=list.New() + queue.PushBack(root) + for queue.Len()>0{ + length:=queue.Len() + for i:=0;i 递归法 + +```go +/** + * Definition for a binary tree node. + * type TreeNode struct { + * Val int + * Left *TreeNode + * Right *TreeNode + * } + */ + var maxDeep int // 全局变量 深度 + var value int //全局变量 最终值 +func findBottomLeftValue(root *TreeNode) int { + if root.Left==nil&&root.Right==nil{//需要提前判断一下（不要这个if的话提交结果会出错，但执行代码不会。防止这种情况出现，故先判断是否只有一个节点） + return root.Val + } + findLeftValue (root,maxDeep) + return value +} +func findLeftValue (root *TreeNode,deep int){ + //最左边的值在左边 + if root.Left==nil&&root.Right==nil{ + if deep>maxDeep{ + value=root.Val + maxDeep=deep + } + } + //递归 + if root.Left!=nil{ + deep++ + findLeftValue(root.Left,deep) + deep--//回溯 + } + if root.Right!=nil{ + deep++ + findLeftValue(root.Right,deep) + deep--//回溯 + } +} +``` + +> 迭代法 + +```go +/** + * Definition for a binary tree node. + * type TreeNode struct { + * Val int + * Left *TreeNode + * Right *TreeNode + * } + */ +func findBottomLeftValue(root *TreeNode) int { + queue:=list.New() + var gradation int + queue.PushBack(root) + for queue.Len()>0{ + length:=queue.Len() + for i:=0;i int: + rows, cols = len(weight), bag_size + 1 + dp = [[0 for _ in range(cols)] for _ in range(rows)] + res = 0 + + # 初始化dp数组. + for i in range(rows): + dp[i][0] = 0 + first_item_weight, first_item_value = weight[0], value[0] + for j in range(1, cols): + if first_item_weight <= j: + dp[0][j] = first_item_value + + # 更新dp数组: 先遍历物品, 再遍历背包. + for i in range(1, len(weight)): + cur_weight, cur_val = weight[i], value[i] + for j in range(1, cols): + if cur_weight > j: # 说明背包装不下当前物品. + dp[i][j] = dp[i - 1][j] # 所以不装当前物品. + else: + # 定义dp数组: dp[i][j] 前i个物品里，放进容量为j的背包，价值总和最大是多少。 + dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - cur_weight]+ cur_val) + if dp[i][j] > res: + res = dp[i][j] + + print(dp) + + +if __name__ == "__main__": + bag_size = 4 + weight = [1, 3, 4] + value = [15, 20, 30] + test_2_wei_bag_problem1(bag_size, weight, value) +``` Go：