添加题567，438，3

Author: xingdali

算法思维/滑动窗口思想.md

@@ -83,3 +83,146 @@ function minWindow(s: string, t: string): string{
 };
 ```
 
+##### [567. 字符串的排列](https://leetcode-cn.com/problems/permutation-in-string/)
+
+给定两个字符串 `s1` 和 `s2`，写一个函数来判断 `s2` 是否包含 `s1` 的排列。换句话说，第一个字符串的排列之一是第二个字符串的 **子串** 。
+
+```tsx
+function checkInclusion(s1:string, s2:string):boolean{
+if (s2.length<s1.length){
+returnfalse
+ }
+const s1arr:number[] =newArray(26).fill(0)
+const s2arr:number[] =newArray(26).fill(0)
+for (let i =0; i<s1.length; i++){
+s1arr[s1.charCodeAt(i) -97]++
+s2arr[s2.charCodeAt(i) -97]++
+ }
+if (s1arr.toString() ===s2arr.toString()){
+returntrue
+ }
+for (let j =s1.length; j<s2.length; j++){
+s2arr[s2.charCodeAt(j) -97]++
+s2arr[s2.charCodeAt(j-s1.length) -97]--
+if (s1arr.toString() ===s2arr.toString()){
+returntrue
+ }
+ }
+returnfalse
+};
+```
+
+```tsx
+function checkInclusion(s1:string, s2:string):boolean{
+const win =newMap<string, number>()
+const need =newMap<string, number>()
+for (let i =0; i<s1.length; i++){
+need.set(s1[i], need.has(s1[i]) ?need.get(s1[i]) +1:1)
+ }
+let left:number=0, right:number=0, match:number=0
+while (right<s2.length){
+const c:string=s2[right]
+right++
+if (need.has(c)){
+win.set(c, win.has(c) ?win.get(c) +1:1)
+if (win.get(c) ===need.get(c)){
+match++
+ }
+ }
+while (right-left>=s1.length){
+if (match===need.size){
+returntrue
+ }
+const d:string=s2[left]
+left++
+if (need.has(d)){
+if (win.get(d) ===need.get(d)){
+match--
+ }
+win.set(d, win.get(d) -1)
+ }
+ }
+ }
+returnfalse
+};
+```
+
+##### [438. 找到字符串中所有字母异位词](https://leetcode-cn.com/problems/find-all-anagrams-in-a-string/)
+
+给定一个字符串 **s** 和一个非空字符串 **p**，找到 **s** 中所有是 **p** 的字母异位词的子串，返回这些子串的起始索引。
+
+```tsx
+function findAnagrams(s:string, p:string):number[]{
+const win =newMap<string, number>()
+const need =newMap<string, number>()
+const ans:number[] = []
+for (let i =0; i<p.length; i++){
+need.set(p[i], need.has(p[i]) ?need.get(p[i]) +1:1)
+ }
+let left:number=0, right:number=0, match:number=0
+while (right<s.length){
+const c:string=s[right]
+right++
+if (need.has(c)){
+win.set(c, win.has(c) ?win.get(c) +1:1)
+if (win.get(c) ===need.get(c)){
+match++
+ }
+ }
+while (right-left>=p.length){
+if (match===need.size){
+ans.push(left)
+ }
+const d:string=s[left]
+left++
+if (need.has(d)){
+if (win.get(d) ===need.get(d)){
+match--
+ }
+win.set(d, win.get(d) -1)
+ }
+ }
+ }
+returnans
+};
+```
+
+##### [3. 无重复字符的最长子串](https://leetcode-cn.com/problems/longest-substring-without-repeating-characters/)
+
+给定一个字符串，请你找出其中不含有重复字符的 **最长子串** 的长度。
+
+```tsx
+function lengthOfLongestSubstring(s:string):number{
+const win =newSet<string>()
+let left:number=0, right:number=0, ans:number=0
+while (right<s.length){
+let c:string=s[right]
+if (win.has(c)){
+c=s[left]
+win.delete(c)
+left++
+ } else{
+win.add(c)
+right++
+ }
+ans=Math.max(ans, win.size)
+ }
+returnans
+};
+```
+
+## 总结
+
+- 和双指针题目类似，更像双指针的升级版，滑动窗口核心点是维护一个窗口集，根据窗口集来进行处理
+- 核心步骤
+- right 右移
+- 收缩
+- left 右移
+- 求结果
+
+## 练习
+
+-[minimum-window-substring](https://leetcode-cn.com/problems/minimum-window-substring/)
+-[permutation-in-string](https://leetcode-cn.com/problems/permutation-in-string/)
+-[find-all-anagrams-in-a-string](https://leetcode-cn.com/problems/find-all-anagrams-in-a-string/)
+-[longest-substring-without-repeating-characters](https://leetcode-cn.com/problems/longest-substring-without-repeating-characters/)