Problem
LeetCode 567. Permutation in String
1. 题目简述
给出两个字符串s1和s2,判断s2中是否含有s1所有字母的一组排列。例如:
Input: s1 = "ab" s2 = "eidbaooo"
Output: True
Explanation: s2 contains one permutation of s1 ("ba").
2. 算法思路
sliding window
这道题和LeetCode-438基本一致,算法也一模一样,只是返回的内容不同,这道题还性对简单点,同样是使用两个数组来记录字母个数,每次滑动窗口时判断是否和目标字母一致。
3. 解法
- 双数组
class Solution {
public boolean checkInclusion(String s1, String s2) {
if (s1.length() > s2.length()) {
return false;
}
int[] dict = new int[26];
int[] window = new int[26];
for (int i = 0; i < s1.length(); i++) {
char c = s1.charAt(i);
dict[c - 'a']++;
}
int start = 0, end = 0;
for (; end < s1.length(); end++) {
char c = s2.charAt(end);
window[c - 'a']++;
}
boolean flag = true;
for (int i = 0; i < 26; i++) {
if (dict[i] != window[i]) {
flag = false;
break;
}
}
if (flag) {
return true;
}
for (start = 1; end < s2.length(); start++, end++) {
window[s2.charAt(start - 1) - 'a']--;
window[s2.charAt(end) - 'a']++;
flag = true;
for (int i = 0; i < 26; i++) {
if (dict[i] != window[i]) {
flag = false;
break;
}
}
if (flag) {
return true;
}
}
return false;
}
}