> For the complete documentation index, see [llms.txt](https://mnunknown.gitbook.io/algorithm-notes/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://mnunknown.gitbook.io/algorithm-notes/two_pointers/615-_two_pointers-_shuang_zhi_zhen_ff0c_chuang_kou.md).

# 双指针，窗口类

* **这两题有一个 trick 和 Minimum Window Substring 非常像，就是维护一个 "curCount" 代表目前 (i,j) 之间 match 上的数量，而通过 hash\[] 的正负充当计数器的作用。**

## [Longest Substring with At Most Two Distinct Characters](https://leetcode.com/problems/longest-substring-with-at-most-two-distinct-characters/)

```java
public class Solution {
    public int lengthOfLongestSubstringTwoDistinct(String s) {
        int maxSize = 0;
        int j = 0;
        int[] hash = new int[256];
        int distinctCount = 0;
        for(int i = 0; i < s.length(); i++){
            while(j < s.length()){
                if(distinctCount == 2 && hash[s.charAt(j)] == 0) break;

                if(hash[s.charAt(j)] == 0) distinctCount ++;
                hash[s.charAt(j++)]++;
            }
            if(j - i > maxSize){
                maxSize = j - i;
            }
            hash[s.charAt(i)]--;
            if(hash[s.charAt(i)] == 0) distinctCount --;
        }

        return maxSize;
    }
}
```

## [Longest Substring with At Most K Distinct Characters](https://github.com/mnmunknown/algorithm-notes/tree/3c03c403dea353c29bf5dc55966e338f3385a7ed/Longest%20Substring%20with%20At%20Most%20K%20Distinct%20Characters/README.md)

和上一题代码完全没有区别，只是把判断条件里面的 count 数字改一下而已。。

```java
public class Solution {
    public int lengthOfLongestSubstringKDistinct(String s, int k) {
        int curCount = 0;
        int j = 0;
        int maxSize = 0;
        int[] hash = new int[256];
        for(int i = 0; i < s.length(); i++){
            while(j < s.length()){
                if(curCount == k && hash[s.charAt(j)] == 0) break;
                if(hash[s.charAt(j)] == 0) curCount ++;
                hash[s.charAt(j++)]++;
            }
            if(j - i > maxSize){
                maxSize = j - i;
            }

            hash[s.charAt(i)]--;
            if(hash[s.charAt(i)] == 0) curCount --;
        }

        return maxSize;
    }
}
```