[LeetCode] 567. Permutation in String 字符串中的全排列

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: False 

Note:

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

给2个字符串s1和s2，写一个函数能够返回是否s1的全排列中存在着一个是s2的子字符串。

虽然题目中有全排列，但跟以前的全排列的题目的解法并不一样，如果遍历s1所有全排列的情况，然后检测其是否为s2的子串，非常不高效。 其实并不需要知道s1的全排列情况，只要知道s2中一个长度和s1一样的子字符串所含的字符一样就可以了。和438. Find All Anagrams in a String 类似。

解法：滑动窗口法

Java:

public class Solution {
    public boolean checkInclusion(String s1, String s2) {
        int len1 = s1.length(), len2 = s2.length();
        if (len1 > len2) return false;

        int[] count = new int[26];
        for (int i = 0; i < len1; i++) {
            count[s1.charAt(i) - 'a']++;
            count[s2.charAt(i) - 'a']--;
        }
        if (allZero(count)) return true;

        for (int i = len1; i < len2; i++) {
            count[s2.charAt(i) - 'a']--;
            count[s2.charAt(i - len1) - 'a']++;
            if (allZero(count)) return true;
        }

        return false;
    }

    private boolean allZero(int[] count) {
        for (int i = 0; i < 26; i++) {
            if (count[i] != 0) return false;
        }
        return true;
    }
}

Java:

public class Solution {
    public boolean checkInclusion(String s1, String s2) {
        int[] map = new int[26];
        int sum = s1.length();
        // construct frequency map
        for(int i = 0; i< s1.length(); i++){
            map[s1.charAt(i) - 'a']++;
        }
        for(int r = 0, l = 0; r < s2.length(); r++){
            char c = s2.charAt(r);
            if(map[c - 'a'] > 0){
                map[c - 'a']--;
                sum--;
                //check for permutation match.
                if(sum == 0) return true;
            }else{
        // if there is enough number for char c or c is never seen before.
        // we move left pointer next to the position where we first saw char c
        // or to the r+1(we never see char c before),
        //and during this process we restore the map.
                while(l<= r && s2.charAt(l) != s2.charAt(r)){
                    map[s2.charAt(l) - 'a'] ++;
                    l++;
                    sum++;
                }
                l++;
            }
        }
        return false;
    }
}　　

Python:

def checkInclusion(self, s1, s2):
    A = [ord(x) - ord('a') for x in s1]
    B = [ord(x) - ord('a') for x in s2]

    target = [0] * 26
    for x in A:
        target[x] += 1

    window = [0] * 26
    for i, x in enumerate(B):
        window[x] += 1
        if i >= len(A):
            window[B[i - len(A)]] -= 1
        if window == target:
            return True
    return False

Python:

class Solution(object):
    def checkInclusion(self, s1, s2):
        """
        :type s1: str
        :type s2: str
        :rtype: bool
        """
        counts = collections.Counter(s1)
        l = len(s1)
        for i in xrange(len(s2)):
            if counts[s2[i]] > 0:
                l -= 1
            counts[s2[i]] -= 1
            if l == 0:
                return True
            start = i + 1 - len(s1)
            if start >= 0:
                counts[s2[start]] += 1
                if counts[s2[start]] > 0:
                    l += 1
        return False　　

C++:　　

class Solution {
public:
    bool checkInclusion(string s1, string s2) {
        int n1 = s1.size(), n2 = s2.size();
        vector<int> m1(128), m2(128);
        for (int i = 0; i < n1; ++i) {
            ++m1[s1[i]]; ++m2[s2[i]];
        }
        if (m1 == m2) return true;
        for (int i = n1; i < n2; ++i) {
            ++m2[s2[i]];
            --m2[s2[i - n1]];
            if (m1 == m2) return true;
        }
        return false;
    }
};　　

C++:

class Solution {
public:
    bool checkInclusion(string s1, string s2) {
        int n1 = s1.size(), n2 = s2.size(), left = 0;
        vector<int> m(128);
        for (char c : s1) ++m[c];
        for (int right = 0; right < n2; ++right) {
            if (--m[s2[right]] < 0) {
                while (++m[s2[left++]] != 0) {}
            } else if (right - left + 1 == n1) return true;
        }
        return n1 == 0;
    }
};

C++:

class Solution {
public:
    bool checkInclusion(string s1, string s2) {
        int n1 = s1.size(), n2 = s2.size(), cnt = n1, left = 0;
        vector<int> m(128);
        for (char c : s1) ++m[c];
        for (int right = 0; right < n2; ++right) {
            if (m[s2[right]]-- > 0) --cnt;
            while (cnt == 0) {
                if (right - left + 1 == n1) return true;
                if (++m[s2[left++]] > 0) ++cnt;
            }
        }
        return false;
    }
};

　　

类似题目：

[LeetCode] 438. Find All Anagrams in a String 找出字符串中所有的变位词

All LeetCode Questions List 题目汇总