[leetcode]438. Find All Anagrams in a String@Java解题报告

Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.The order of output does not matter.Example 1:Input:
s: "cbaebabacd" p: "abc"

Output:
[0, 6]

Explanation:
The substring with start index = 0 is "cba", which is an anagram of "abc".
The substring with start index = 6 is "bac", which is an anagram of "abc".
Example 2:Input:
s: "abab" p: "ab"

Output:
[0, 1, 2]

Explanation:
The substring with start index = 0 is "ab", which is an anagram of "ab".
The substring with start index = 1 is "ba", which is an anagram of "ab".
The substring with start index = 2 is "ab", which is an anagram of "ab".


package go.jacob.day0304.array.滑动窗口;

import java.util.ArrayList;
import java.util.List;

/**
* Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.

Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.

The order of output does not matter.

Example 1:

Input:
s: "cbaebabacd" p: "abc"

Output:
[0, 6]

Explanation:
The substring with start index = 0 is "cba", which is an anagram of "abc".
The substring with start index = 6 is "bac", which is an anagram of "abc".
Example 2:

Input:
s: "abab" p: "ab"

Output:
[0, 1, 2]

Explanation:
The substring with start index = 0 is "ab"
4000
, which is an anagram of "ab".
The substring with start index = 1 is "ba", which is an anagram of "ab".
The substring with start index = 2 is "ab", which is an anagram of "ab".
*/
public class P438_FindAllAnagramsInAString {

public List<Integer> findAnagrams(String s, String p) {
List<Integer> res = new ArrayList<Integer>();

if (s == null || s.length() == 0 || p == null || p.length() == 0)
return res;

int l = 0, r = 0;
int count = p.length();

int[] hash = new int[256];

for (char c : p.toCharArray()) {
hash[c]++;
}

while (r < s.length()) {

if (hash[s.charAt(r++)]-- > 0) {
count--;
}

if (count == 0)
res.add(l);

if (r - l == p.length() && hash[s.charAt(l++)]++ >= 0)//要增加l前，需要判断
count++;
}

return res;

}

public static void main(String[] args) {
int a=6;
if(a++>6)
;
System.out.println(a);

}
}