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Merge branch 'youngyangyang04:master' into master
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@ -894,7 +894,58 @@ var strStr = function (haystack, needle) {
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};
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```
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Swift 版本
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> 前缀表统一减一
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```swift
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func strStr(_ haystack: String, _ needle: String) -> Int {
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let s = Array(haystack), p = Array(needle)
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guard p.count != 0 else { return 0 }
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// 2 pointer
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var j = -1
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var next = [Int](repeating: -1, count: needle.count)
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// KMP
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getNext(&next, needle: p)
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for i in 0 ..< s.count {
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while j >= 0 && s[i] != p[j + 1] {
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//不匹配之后寻找之前匹配的位置
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j = next[j]
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}
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if s[i] == p[j + 1] {
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//匹配,双指针同时后移
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j += 1
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}
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if j == (p.count - 1) {
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//出现匹配字符串
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return i - p.count + 1
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}
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}
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return -1
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}
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//前缀表统一减一
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func getNext(_ next: inout [Int], needle: [Character]) {
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var j: Int = -1
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next[0] = j
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// i 从 1 开始
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for i in 1 ..< needle.count {
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while j >= 0 && needle[i] != needle[j + 1] {
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j = next[j]
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}
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if needle[i] == needle[j + 1] {
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j += 1;
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}
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next[i] = j
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}
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print(next)
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}
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```
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-----------------------
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
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@ -405,7 +405,7 @@ class Solution:
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if len(path) == k:
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res.append(path[:])
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return
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for i in range(StartIndex, n-(k-len(path)) + 2):
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for i in range(StartIndex, n + 1):
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path.append(i)
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backtrack(n, k, i+1)
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path.pop()
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@ -414,7 +414,7 @@ class Solution:
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```
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剪枝:
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```python3
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```python
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class Solution:
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def combine(self, n: int, k: int) -> List[List[int]]:
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res=[] #存放符合条件结果的集合
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@ -423,7 +423,7 @@ class Solution:
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if len(path) == k:
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res.append(path[:])
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return
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for i in range(startIndex,n-(k-len(path))+2): #优化的地方
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for i in range(startIndex,n - (k - len(path)) + 2): #优化的地方
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path.append(i) #处理节点
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backtrack(n,k,i+1) #递归
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path.pop() #回溯,撤销处理的节点
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@ -409,5 +409,56 @@ int* postorderTraversal(struct TreeNode* root, int* returnSize){
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}
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```
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Swift:
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前序遍历:(144.二叉树的前序遍历)
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```Swift
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func preorderTraversal(_ root: TreeNode?) -> [Int] {
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var res = [Int]()
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preorder(root, res: &res)
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return res
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}
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func preorder(_ root: TreeNode?, res: inout [Int]) {
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if root == nil {
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return
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}
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res.append(root!.val)
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preorder(root!.left, res: &res)
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preorder(root!.right, res: &res)
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}
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```
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中序遍历:(94. 二叉树的中序遍历)
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```Swift
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func inorderTraversal(_ root: TreeNode?) -> [Int] {
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var res = [Int]()
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inorder(root, res: &res)
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return res
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}
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func inorder(_ root: TreeNode?, res: inout [Int]) {
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if root == nil {
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return
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}
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inorder(root!.left, res: &res)
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res.append(root!.val)
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inorder(root!.right, res: &res)
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}
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```
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后序遍历:(145. 二叉树的后序遍历)
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```Swift
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func postorderTraversal(_ root: TreeNode?) -> [Int] {
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var res = [Int]()
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postorder(root, res: &res)
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return res
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}
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func postorder(_ root: TreeNode?, res: inout [Int]) {
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if root == nil {
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return
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}
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postorder(root!.left, res: &res)
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postorder(root!.right, res: &res)
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res.append(root!.val)
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}
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```
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-----------------------
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<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
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