diff --git a/problems/0055.跳跃游戏.md b/problems/0055.跳跃游戏.md index a25c831a..ae2cbd1e 100644 --- a/problems/0055.跳跃游戏.md +++ b/problems/0055.跳跃游戏.md @@ -122,6 +122,24 @@ class Solution: ``` Go： +```Go +func canJUmp(nums []int) bool { + if len(nums)<=1{ + return true + } + dp:=make([]bool,len(nums)) + dp[0]=true + for i:=1;i=0;j--{ + if dp[j]&&nums[j]+j>=i{ + dp[i]=true + break + } + } + } + return dp[len(nums)-1] +} +``` 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 01b6c89c..5605e13d 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 @@ -302,6 +302,53 @@ class Solution: Go： +JavaScript: + +递归法： + +```javascript +/** + * @param {TreeNode} root + * @return {number} + */ +var minDepth1 = function(root) { + if(!root) return 0; + // 到叶子节点 返回 1 + if(!root.left && !root.right) return 1; + // 只有右节点时 递归右节点 + if(!root.left) return 1 + minDepth(root.right);、 + // 只有左节点时 递归左节点 + if(!root.right) return 1 + minDepth(root.left); + return Math.min(minDepth(root.left), minDepth(root.right)) + 1; +}; +``` + +迭代法： + +```javascript +/** +* @param {TreeNode} root +* @return {number} +*/ +var minDepth = function(root) { + if(!root) return 0; + const queue = [root]; + let dep = 0; + while(true) { + let size = queue.length; + dep++; + while(size--){ + const node = queue.shift(); + // 到第一个叶子节点 返回 当前深度 + if(!node.left && !node.right) return dep; + node.left && queue.push(node.left); + node.right && queue.push(node.right); + } + } +}; +``` + + ----------------------- diff --git a/problems/0383.赎金信.md b/problems/0383.赎金信.md index 2f7ae4d8..527945fa 100644 --- a/problems/0383.赎金信.md +++ b/problems/0383.赎金信.md @@ -136,7 +136,34 @@ class Solution { ``` Python： - +``` +class Solution(object): + def canConstruct(self, ransomNote, magazine): + """ + :type ransomNote: str + :type magazine: str + :rtype: bool + """ + + # use a dict to store the number of letter occurance in ransomNote + hashmap = dict() + for s in ransomNote: + if s in hashmap: + hashmap[s] += 1 + else: + hashmap[s] = 1 + + # check if the letter we need can be found in magazine + for l in magazine: + if l in hashmap: + hashmap[l] -= 1 + + for key in hashmap: + if hashmap[key] > 0: + return False + + return True +``` Go： diff --git a/problems/0454.四数相加II.md b/problems/0454.四数相加II.md index a2ef928b..ad928a3f 100644 --- a/problems/0454.四数相加II.md +++ b/problems/0454.四数相加II.md @@ -121,6 +121,37 @@ class Solution { Python： +``` +class Solution(object): + def fourSumCount(self, nums1, nums2, nums3, nums4): + """ + :type nums1: List[int] + :type nums2: List[int] + :type nums3: List[int] + :type nums4: List[int] + :rtype: int + """ + # use a dict to store the elements in nums1 and nums2 and their sum + hashmap = dict() + for n1 in nums1: + for n2 in nums2: + if n1 + n2 in hashmap: + hashmap[n1+n2] += 1 + else: + hashmap[n1+n2] = 1 + + # if the -(a+b) exists in nums3 and nums4, we shall add the count + count = 0 + for n3 in nums3: + for n4 in nums4: + key = - n3 - n4 + if key in hashmap: + count += hashmap[key] + return count + + +``` + Go： diff --git a/problems/0516.最长回文子序列.md b/problems/0516.最长回文子序列.md index d2253a4e..7736b69f 100644 --- a/problems/0516.最长回文子序列.md +++ b/problems/0516.最长回文子序列.md @@ -175,36 +175,35 @@ Python： Go： ```Go func longestPalindromeSubseq(s string) int { - str:=[]byte(s) - dp:=make([][]int,len(s)) - for i:=0;i=0;j--{ - if str[i]==str[j]{ - if j==i-1{ - dp[j][i]=2 - }else{ - dp[j][i]=dp[j+1][i-1]+2 - } - }else{ - dp[j][i]=Max(dp[j+1][i],dp[j][i-1]) - } - } - } - return dp[0][len(s)-1] -} -func Max(a,b int)int{ - if a>b{ - return a - } - return b + lenth:=len(s) + dp:=make([][]int,lenth) + for i:=0;i=0;i--{ + for j:=i+1;j int: + res = [] + r = float("inf") + def buildaList(root): //把二叉搜索树转换成有序数组 + if not root: return None + if root.left: buildaList(root.left) //左 + res.append(root.val) //中 + if root.right: buildaList(root.right) //右 + return res + + buildaList(root) + for i in range(len(res)-1): // 统计有序数组的最小差值 + r = min(abs(res[i]-res[i+1]),r) + return r +``` Go： @@ -188,4 +209,4 @@ 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/二叉树的迭代遍历.md b/problems/二叉树的迭代遍历.md index 2647616b..594d498f 100644 --- a/problems/二叉树的迭代遍历.md +++ b/problems/二叉树的迭代遍历.md @@ -160,11 +160,175 @@ Java： Python： +```python3 +# 前序遍历-迭代-LC144_二叉树的前序遍历 +class Solution: + def preorderTraversal(self, root: TreeNode) -> List[int]: + # 根结点为空则返回空列表 + if not root: + return [] + stack = [root] + result = [] + while stack: + node = stack.pop() + # 中结点先处理 + result.append(node.val) + # 右孩子先入栈 + if node.right: + stack.append(node.right) + # 左孩子后入栈 + if node.left: + stack.append(node.left) + return result + +# 中序遍历-迭代-LC94_二叉树的中序遍历 +class Solution: + def inorderTraversal(self, root: TreeNode) -> List[int]: + if not root: + return [] + stack = [] # 不能提前将root结点加入stack中 + result = [] + cur = root + while cur or stack: + # 先迭代访问最底层的左子树结点 + if cur: + stack.append(cur) + cur = cur.left + # 到达最左结点后处理栈顶结点 + else: + cur = stack.pop() + result.append(cur.val) + # 取栈顶元素右结点 + cur = cur.right + return result + +# 后序遍历-迭代-LC145_二叉树的后序遍历 +class Solution: + def postorderTraversal(self, root: TreeNode) -> List[int]: + if not root: + return [] + stack = [root] + result = [] + while stack: + node = stack.pop() + # 中结点先处理 + result.append(node.val) + # 左孩子先入栈 + if node.left: + stack.append(node.left) + # 右孩子后入栈 + if node.right: + stack.append(node.right) + # 将最终的数组翻转 + return result[::-1] +``` Go： +> 迭代法前序遍历 +```go +//迭代法前序遍历 +/** + type Element struct { + // 元素保管的值 + Value interface{} + // 内含隐藏或非导出字段 +} +func (l *List) Back() *Element +前序遍历：中左右 +压栈顺序：右左中 + **/ +func preorderTraversal(root *TreeNode) []int { + if root == nil { + return nil + } + var stack = list.New() + stack.PushBack(root.Right) + stack.PushBack(root.Left) + res:=[]int{} + res=append(res,root.Val) + for stack.Len()>0 { + e:=stack.Back() + stack.Remove(e) + node := e.Value.(*TreeNode)//e是Element类型，其值为e.Value.由于Value为接口，所以要断言 + if node==nil{ + continue + } + res=append(res,node.Val) + stack.PushBack(node.Right) + stack.PushBack(node.Left) + } + return res +} +``` + +> 迭代法后序遍历 + +```go +//迭代法后序遍历 +//后续遍历：左右中 +//压栈顺序：中右左(按照前序遍历思路)，再反转结果数组 +func postorderTraversal(root *TreeNode) []int { + if root == nil { + return nil + } + var stack = list.New() + stack.PushBack(root.Left) + stack.PushBack(root.Right) + res:=[]int{} + res=append(res,root.Val) + for stack.Len()>0 { + e:=stack.Back() + stack.Remove(e) + node := e.Value.(*TreeNode)//e是Element类型，其值为e.Value.由于Value为接口，所以要断言 + if node==nil{ + continue + } + res=append(res,node.Val) + stack.PushBack(node.Left) + stack.PushBack(node.Right) + } + for i:=0;i 迭代法中序遍历 + +```go +//迭代法中序遍历 +func inorderTraversal(root *TreeNode) []int { + rootRes:=[]int{} + if root==nil{ + return nil + } + stack:=list.New() + node:=root + //先将所有左节点找到，加入栈中 + for node!=nil{ + stack.PushBack(node) + node=node.Left + } + //其次对栈中的每个节点先弹出加入到结果集中，再找到该节点的右节点的所有左节点加入栈中 + for stack.Len()>0{ + e:=stack.Back() + node:=e.Value.(*TreeNode) + stack.Remove(e) + //找到该节点的右节点，再搜索他的所有左节点加入栈中 + rootRes=append(rootRes,node.Val) + node=node.Right + for node!=nil{ + stack.PushBack(node) + node=node.Left + } + } + return rootRes +} +``` -----------------------