mirror of
https://github.com/youngyangyang04/leetcode-master.git
synced 2025-07-08 16:54:50 +08:00
Merge branch 'youngyangyang04:master' into master
This commit is contained in:
@ -131,7 +131,20 @@ class Solution {
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```
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Python:
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```python
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class Solution:
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def rob(self, nums: List[int]) -> int:
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if len(nums) == 0:
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return 0
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if len(nums) == 1:
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return nums[0]
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dp = [0] * len(nums)
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dp[0] = nums[0]
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dp[1] = max(nums[0], nums[1])
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for i in range(2, len(nums)):
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dp[i] = max(dp[i-2]+nums[i], dp[i-1])
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return dp[-1]
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```
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Go:
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```Go
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@ -308,6 +308,35 @@ class Solution:
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Go:
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递归版本
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```go
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/**
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* Definition for a binary tree node.
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* type TreeNode struct {
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* Val int
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* Left *TreeNode
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* Right *TreeNode
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* }
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*/
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//本题直接就是求有多少个节点,无脑存进数组算长度就行了。
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func countNodes(root *TreeNode) int {
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if root == nil {
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return 0
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}
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res := 1
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if root.Right != nil {
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res += countNodes(root.Right)
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}
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if root.Left != nil {
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res += countNodes(root.Left)
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}
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return res
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}
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```
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JavaScript:
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递归版本
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@ -287,6 +287,25 @@ class Solution {
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Python:
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> 动态规划
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```python
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class Solution:
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def rob(self, root: TreeNode) -> int:
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result = self.robTree(root)
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return max(result[0], result[1])
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#长度为2的数组,0:不偷,1:偷
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def robTree(self, cur):
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if not cur:
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return (0, 0) #这里返回tuple, 也可以返回list
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left = self.robTree(cur.left)
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right = self.robTree(cur.right)
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#偷cur
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val1 = cur.val + left[0] + right[0]
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#不偷cur
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val2 = max(left[0], left[1]) + max(right[0], right[1])
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return (val2, val1)
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```
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Go:
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@ -206,7 +206,43 @@ class Solution:
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```
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Go:
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```go
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func findMaxForm(strs []string, m int, n int) int {
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// 定义数组
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dp := make([][]int, m+1)
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for i,_ := range dp {
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dp[i] = make([]int, n+1 )
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}
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// 遍历
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for i:=0;i<len(strs);i++ {
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zeroNum,oneNum := 0 , 0
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//计算0,1 个数
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//或者直接strings.Count(strs[i],"0")
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for _,v := range strs[i] {
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if v == '0' {
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zeroNum++
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}
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}
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oneNum = len(strs[i])-zeroNum
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// 从后往前 遍历背包容量
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for j:= m ; j >= zeroNum;j-- {
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for k:=n ; k >= oneNum;k-- {
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// 推导公式
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dp[j][k] = max(dp[j][k],dp[j-zeroNum][k-oneNum]+1)
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}
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}
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//fmt.Println(dp)
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}
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return dp[m][n]
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}
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func max(a,b int) int {
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if a > b {
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return a
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}
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return b
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}
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```
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@ -151,7 +151,30 @@ public:
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Java:
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递归
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```java
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class Solution {
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TreeNode pre;// 记录上一个遍历的结点
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int result = Integer.MAX_VALUE;
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public int getMinimumDifference(TreeNode root) {
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if(root==null)return 0;
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traversal(root);
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return result;
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}
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public void traversal(TreeNode root){
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if(root==null)return;
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//左
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traversal(root.left);
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//中
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if(pre!=null){
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result = Math.min(result,root.val-pre.val);
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}
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pre = root;
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//右
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traversal(root.right);
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}
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}
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```
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```Java
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class Solution {
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TreeNode pre;// 记录上一个遍历的结点
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@ -257,6 +257,83 @@ Java:
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Python:
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100.相同的树
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> 递归法
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```python
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class Solution:
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def isSameTree(self, p: TreeNode, q: TreeNode) -> bool:
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return self.compare(p, q)
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def compare(self, tree1, tree2):
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if not tree1 and tree2:
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return False
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elif tree1 and not tree2:
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return False
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elif not tree1 and not tree2:
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return True
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elif tree1.val != tree2.val: #注意这里我没有使用else
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return False
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#此时就是:左右节点都不为空,且数值相同的情况
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#此时才做递归,做下一层的判断
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compareLeft = self.compare(tree1.left, tree2.left) #左子树:左、 右子树:左
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compareRight = self.compare(tree1.right, tree2.right) #左子树:右、 右子树:右
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isSame = compareLeft and compareRight #左子树:中、 右子树:中(逻辑处理)
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return isSame
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```
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257.二叉的所有路径
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> 递归中隐藏着回溯
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```python
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class Solution:
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def binaryTreePaths(self, root: TreeNode) -> List[str]:
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result = []
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path = []
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if not root:
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return result
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self.traversal(root, path, result)
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return result
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def traversal(self, cur, path, result):
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path.append(cur.val)
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#这才到了叶子节点
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if not cur.left and not cur.right:
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sPath = ""
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for i in range(len(path)-1):
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sPath += str(path[i])
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sPath += "->"
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sPath += str(path[len(path)-1])
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result.append(sPath)
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return
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if cur.left:
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self.traversal(cur.left, path, result)
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path.pop() #回溯
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if cur.right:
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self.traversal(cur.right, path, result)
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path.pop() #回溯
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```
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> 精简版
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```python
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class Solution:
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def binaryTreePaths(self, root: TreeNode) -> List[str]:
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result = []
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path = ""
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if not root:
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return result
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self.traversal(root, path, result)
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return result
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def traversal(self, cur, path, result):
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path += str(cur.val) #中
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if not cur.left and not cur.right:
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result.append(path)
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return
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if cur.left:
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self.traversal(cur.left, path+"->", result) #左 回溯就隐藏在这里
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if cur.right:
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self.traversal(cur.right, path+"->", result) #右 回溯就隐藏在这里
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```
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Go:
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@ -273,7 +273,40 @@ Python:
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Go:
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```go
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func test_2_wei_bag_problem1(weight, value []int, bagWeight int) int {
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// 定义dp数组
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dp := make([][]int, len(weight))
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for i, _ := range dp {
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dp[i] = make([]int, bagWeight+1)
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}
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// 初始化
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for j := bagWeight; j >= weight[0]; j-- {
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dp[0][j] = dp[0][j-weight[0]] + value[0]
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}
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// 递推公式
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for i := 1; i < len(weight); i++ {
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//正序,也可以倒序
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for j := weight[i];j<= bagWeight ; j++ {
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dp[i][j] = max(dp[i-1][j], dp[i-1][j-weight[i]]+value[i])
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}
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}
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return dp[len(weight)-1][bagWeight]
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}
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func max(a,b int) int {
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if a > b {
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return a
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}
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return b
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}
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func main() {
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weight := []int{1,3,4}
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value := []int{15,20,30}
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test_2_wei_bag_problem1(weight,value,4)
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}
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```
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-----------------------
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* 作者微信:[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw)
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@ -219,7 +219,36 @@ Python:
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Go:
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```go
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func test_1_wei_bag_problem(weight, value []int, bagWeight int) int {
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// 定义 and 初始化
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dp := make([]int,bagWeight+1)
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// 递推顺序
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for i := 0 ;i < len(weight) ; i++ {
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// 这里必须倒序,区别二维,因为二维dp保存了i的状态
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for j:= bagWeight; j >= weight[i] ; j-- {
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// 递推公式
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dp[j] = max(dp[j], dp[j-weight[i]]+value[i])
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}
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}
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//fmt.Println(dp)
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return dp[bagWeight]
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}
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func max(a,b int) int {
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if a > b {
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return a
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}
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return b
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}
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func main() {
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weight := []int{1,3,4}
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value := []int{15,20,30}
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test_1_wei_bag_problem(weight,value,4)
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}
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```
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Reference in New Issue
Block a user