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567. Permutation in String
Given two strings s1 and s2, write a function to return true if s2 contains the permutation of s1. In other words, one of the first string’s permutations is the substring of the second string.
Example 1:
Input: s1 = "ab" s2 = "eidbaooo"
Output: True
Explanation: s2 contains one permutation of s1 ("ba").Example 2:
Input:s1= "ab" s2 = "eidboaoo"
Output: FalseNote:
-
The input strings only contain lower case letters.
-
The length of both given strings is in range [1, 10,000].
解题分析
滑动窗口大法好！
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/**
* Runtime: 14 ms, faster than 45.41% of Java online submissions for Permutation in String.
* Memory Usage: 39.4 MB, less than 7.69% of Java online submissions for Permutation in String.
*/
public boolean checkInclusion(String s1, String s2) {
if (Objects.isNull(s1) || s1.length() == 0) {
return true;
}
if (Objects.isNull(s2) || s2.length() == 0 || s2.length() < s1.length()) {
return false;
}
Map<Character, Integer> needs = new HashMap<>();
for (char c : s1.toCharArray()) {
needs.put(c, needs.getOrDefault(c, 0) + 1);
}
int left = 0, right = 0;
int match = 0;
Map<Character, Integer> windows = new HashMap<>();
while (right < s2.length()) {
char rChar = s2.charAt(right);
if (needs.containsKey(rChar)) {
int rCount = windows.getOrDefault(rChar, 0) + 1;
windows.put(rChar, rCount);
if (rCount == needs.get(rChar)) {
match++;
}
}
right++;
while (match == needs.size()) {
if (right - left == s1.length()) {
return true;
}
char lChar = s2.charAt(left);
if (needs.containsKey(lChar)) {
int lCount = windows.get(lChar) - 1;
windows.put(lChar, lCount);
if (lCount < needs.get(lChar)) {
match--;
}
}
left++;
}
}
return false;
}
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/**
* 自己实现
*/
public boolean checkInclusion(String s1, String s2) {
if (s1 == null || s1.isEmpty()
|| s2 == null || s2.isEmpty()
|| s2.length() < s1.length()) {
return false;
}
Map<Character, Integer> need = new HashMap<>();
for (char c : s1.toCharArray()) {
need.put(c, need.getOrDefault(c, 0) + 1);
}
Map<Character, Integer> window = new HashMap<>();
int left = 0, right = 0;
int valid = 0;
while (right < s2.length()) {
char rc = s2.charAt(right);
right++;
if (need.containsKey(rc)) {
window.put(rc, window.getOrDefault(rc, 0) + 1);
if (window.get(rc).equals(need.get(rc))) {
valid++;
}
}
while (need.size() == valid) {
if (right - left == s1.length()) {
return true;
}
char lc = s2.charAt(left);
if (window.containsKey(lc)) {
window.put(lc, window.get(lc) - 1);
if (window.get(lc) < need.get(lc)) {
valid--;
}
}
left++;
}
}
return false;
}