Leetcode:567.字符串的排列

题目: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。

示例1:

输入: s1 = "ab" s2 = "eidbaooo"
输出: True
解释: s2 包含 s1 的排列之一 ("ba").
 

示例2:

输入: s1= "ab" s2 = "eidboaoo"
输出: False

AC代码,滑动窗口算法:

class Solution {
    public boolean checkInclusion(String s1, String s2) {
        int i = 0;
        int j = 0;
        int n_s1 = s1.length();
        int n_s2 = s2.length();

        int[] counts = new int[26];
        for(int x = 0;x < n_s1;++x){
            ++counts[s1.charAt(x) - 97];
        }

        int[] temp = new int[26];
        int[] index_count = new int[26];
        int hasReturn = 0;

        while(i <= n_s2 - n_s1 && j < s2.length()){
            if(counts[s2.charAt(j) - 97] == 0){
                i = ++j;
                temp = new int[26];
                index_count = new int[26];
                hasReturn = 0;
            }
            else{
                int j_index = s2.charAt(j) - 97;
                int i_index = s2.charAt(i) - 97;
                if(temp[j_index] + 1 >counts[j_index]){
                    --temp[i_index];
                    if(temp[i_index] != counts[i_index]){
                        if(index_count[i_index] == 1){
                            index_count[i_index] = 0;
                            hasReturn -= counts[i_index];
                        }
                    }
                    ++i;

                }
                else if(temp[j_index] + 1 == counts[j_index]){
                    index_count[j_index] = 1;
                    hasReturn += counts[j_index];
                    if(hasReturn == n_s1)
                        return true;
                    ++temp[j_index];
                    ++j;
                }
                else{
                    ++temp[j_index];
                    ++j;
                }
            }
        }
        return false;
    }
}

解法2:暴力求解

 

来源：CSDN

作者：liuliuliudy

链接：https://blog.csdn.net/qq_33607393/article/details/104539254