滑动窗口题目总结

 转自：https://leetcode-cn.com/problems/minimum-window-substring/solution/hua-dong-chuang-kou-suan-fa-tong-yong-si-xiang-by-/

3. 无重复字符的最长子串

最典型的滑动窗口题目

right要找到新字符，才加1，扩大窗口
left要发现旧字符就加1，缩小窗口，直到right处的flag[s[right]]值变为0，right才继续增加

需要注意的就是right++;的位置，需要考虑清楚

#define MAX(x,y) (x)>(y)?(x):(y)
int lengthOfLongestSubstring(char * s){
    int left=0,right=0;
    int len=strlen(s);
    int maxlen=0;
    int flag[256]={0};
    while(right

76. 最小覆盖子串 

char * minWindow(char * s, char * t){
	int need[300] = { 0 };
	int window[300] = { 0 };
	int needlen = 0,minlen=INT_MAX;
	char* res = malloc(100000);
	memset(res, '\0', 100000);
	for (int i = 0; i < strlen(t); i++) {
		char idx = t[i];
		if (need[idx] == 0)
			needlen++;
		need[idx]++;
	}
	int match = 0;
	int right = 0, left = 0, start = 0;
	while (right < strlen(s)) {
		char c1 = s[right];
		if (need[c1]) {
			window[c1]++;
			if (window[c1] == need[c1]) {
				match++;
			}
		}

		right++;
		while (match == needlen) {
			if (right - left < minlen) {
				start = left;
				minlen = right - left;
			}
			char c2 = s[left];
			if (need[c2]) {
				window[c2]--;
				if (window[c2] < need[c2]) {
					match--;
				}
			}
			left++;
		}
	}
	if (minlen == INT_MAX)res = "";
	else {
		res = memcpy(res, s + start, minlen);
		
	}
	return res;
}

438. 找到字符串中所有字母异位词

与上一题76. 最小覆盖子串 的区别在下面这两句，因为要找长度完全符合只是顺序不同的字符串。

while (match == needlen) {
            if (right - left == strlen(p)) {
                res[k++]=left;
            }

/**
 * Note: The returned array must be malloced, assume caller calls free().
 */
int* findAnagrams(char * s, char * p, int* returnSize){
	int need[200] = { 0 };
	int window[200] = { 0 };
	int needlen = 0, minlen = INT_MAX;
	int* res = malloc(1000000);
	for (int i = 0; i < strlen(p); i++) {
		char idx = p[i];
		if (need[idx] == 0)
			needlen++;
		need[idx]++;
	}
	int match = 0,k=0;
	int right = 0, left = 0;
	while (right < strlen(s)) {
		char c1 = s[right];
		if (need[c1]) {
			window[c1]++;
			if (window[c1] == need[c1]) {
				match++;
			}
		}
		right++;
		while (match == needlen) {
			if (right - left == strlen(p)) {
				res[k++]=left;
			}
			char c2 = s[left];
			if (need[c2]) {
				window[c2]--;
				if (window[c2] < need[c2]) {
					match--;
				}
			}
			left++;
		}
	}
	*returnSize = k;
	return res;
}

最后总结
通过上面三道题，我们可以总结出滑动窗口算法的抽象思想：

int left = 0, right = 0;

while (right < s.size()) {
    window.add(s[right]);
    right++;
    
    while (valid) {
        window.remove(s[left]);
        left++;
    }
}