题目描述
思路:滑动窗口模板
- 定义需要维护的变量
Map<Character, Integer> map = new HashMap<>();
Map<Character, Integer> map_s1 = new HashMap<>();
for (char c : s1.toCharArray()) {
map_s1.put(c, map_s1.getOrDefault(c, 0) + 1);
}
- 根据题意可知:窗口为固定大小所以用if
if (end - start + 1 == s1.length()) {
char startChar = s2.charAt(start);
map.put(startChar, map.get(startChar) - 1);
if (map.get(startChar) == 0) {
map.remove(startChar);
}
start ++;
}
方法一:
class Solution {
public boolean checkInclusion(String s1, String s2) {
// 1. 定义需要维护的变量
Map<Character, Integer> map = new HashMap<>();
Map<Character, Integer> map_s1 = new HashMap<>();
for (char c : s1.toCharArray()) {
map_s1.put(c, map_s1.getOrDefault(c, 0) + 1);
}
// 2. 定义窗口边界
int start = 0;
for (int end = 0; end < s2.length(); end ++) {
// 3. 更新需要维护的变量
char currentChar = s2.charAt(end);
map.put(currentChar, map.getOrDefault(currentChar, 0) + 1);
if (map.equals(map_s1)) {
return true;
}
if (end - start + 1 == s1.length()) {
char startChar = s2.charAt(start);
map.put(startChar, map.get(startChar) - 1);
if (map.get(startChar) == 0) {
map.remove(startChar);
}
start ++;
}
}
return false;
}
}
标签:map,排列,LeetCode567,startChar,s1,char,start,字符串,end
From: https://www.cnblogs.com/keyongkang/p/17874907.html