LCR 014.字符串的排列

文章目录

• 题意
• 思路
• 代码

题意

题目链接

思路

滑动窗口

代码

class Solution {
public:
    bool judge(int a[], int b[]) {
        for (int i = 0; i < 26; i++)
            if (a[i] != b[i])
                return false;
        return true;
    }
    bool checkInclusion(string s1, string s2) {
        int a[26] = {0};
        int b[26] = {0};
        if (s1.size() > s2.size())
            return false;
        for (int i = 0; i < s1.size(); i++) {
            a[s1[i] - 'a'] ++;
            b[s2[i] - 'a'] ++;
        }
        
        if (judge(a, b)) {
            return true;
        }
        for (int i = 0, j = s1.size(); j < s2.size(); i++, j++) {
            b[s2[i] - 'a']--;
            b[s2[j] - 'a']++;
            if (judge(a,b))
                return true;
        }
        return false;
    }
};