LeetCode[567]Permutation in String(Java)

Description:

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:TrueExplanation: s2 contains one permutation of s1 ("ba").

Example 2:

Input:s1= "ab" s2 = "eidboaoo"Output: False

Note:

1. The input strings only contain lower case letters.
2. The length of both given strings is in range [1, 10,000].

Solution:

滑窗法
每次s2中截取和s1等长的字符串，与s1进行比较，利用数组来统计a-z的数量，辅助比较。

class Solution {    public boolean checkInclusion(String s1, String s2) {        if(s1.length() > s2.length()){            return false;        }        int[] chars1 = new int[26];        for(int i = 0; i < s1.length(); i++){            chars1[s1.charAt(i) - 'a'] ++;        }        for(int i = 0; i <= s2.length() - s1.length(); i++){            String target = s2.substring(i, i+s1.length());            if(helper(chars1, target)){                return true;            }        }        return false;    }    public boolean helper(int[] chars1, String target){        int[] temp = new int[26];        for(int i = 0; i < chars1.length; i++){            temp[i] = chars1[i];        }        for(int i = 0; i < target.length(); i++){            if(temp[target.charAt(i) - 'a'] > 0){                temp[target.charAt(i) - 'a'] --;            }else{                return false;            }        }        return true;    }}

优化

class Solution {    public boolean checkInclusion(String s1, String s2) {        if(s1.length() > s2.length()){            return false;        }        int[] count = new int[26];        for(int i = 0; i < s1.length(); i++){            count[s1.charAt(i) - 'a'] ++;            count[s2.charAt(i) - 'a'] --;        }                if(helper(count)){            return true;        }                for(int i = s1.length(); i < s2.length(); i++){            count[s2.charAt(i) - 'a'] --;            count[s2.charAt(i - s1.length()) - 'a'] ++;            if(helper(count)){                return true;            }        }        return false;            }    public boolean helper(int[] count){        for(int num : count){            if(num != 0){                return false;            }        }        return true;    }}