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: False

 

Note:

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

 

---

 

Naive解法：从S2左边开始，找到所有可能的长度和S1相同的子字符串，然后sort该子字符串，如果和S1的sort后的字符串一样，就返回true。

class Solution {
public:
    bool checkInclusion(string s1, string s2) {
        if(s2.size()<s1.size()){
            return false;
        }
        string target = s1;
        sort(target.begin(),target.end());
        const int SIZE = s1.size();
        for(int i=0;i<s2.size()-SIZE+1;i++){
            string tmp = s2.substr(i,SIZE);
            sort(tmp.begin(),tmp.end());
            if(tmp==target){
                return true;
            }
        }
        return false;
    }
};

效率太低，时间复杂度是O(N*M*log(M))，没有合理利用已经计算过的信息：e.g., 前后两个连续的子字符串，s_i,....s_j 和s_(i+1),....s_(j+1)，第二个相比第一个多了s_(j+1)而少了s_i，中间的信息重复计算了。

 

解法二：

class Solution {
public:
    bool checkInclusion(string s1, string s2) {
        vector<int> freq(26,0);
        int count = 0;
        for(char c: s1){
            if(freq[c-'a']==0){
                count++;
            }
            freq[c-'a']++;
        }
        
        for(int i=0;i<s1.size();i++){
            if(freq[s2[i]-'a']==0){
                count++;
            }
            freq[s2[i]-'a']--;
            if(freq[s2[i]-'a']==0){
                count--;
            }
        }
        
        if(count==0) return true;
        
        for(int i=s1.size();i<s2.size();i++){
            if(freq[s2[i]-'a']==0){
                count++;
            }
            freq[s2[i]-'a']--;
            if(freq[s2[i]-'a']==0){
                count--;
            }
            
            if(freq[s2[i-s1.size()]-'a']==0){
                count++;
            }
            freq[s2[i-s1.size()]-'a']++;
            if(freq[s2[i-s1.size()]-'a']==0){
                count--;
            }
            
            if(count==0) return true;
        }
        return false;
    }
};

 

grandyang 解法一：两个字符串分别用两个hash table来记录，直接比较hash table。

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;
    }
};

 

另一解法：

class Solution {
public:
    bool checkInclusion(string s1, string s2) {
        if(s1.size()>s2.size()){
            return false;
        }
        unordered_map<char,int> freq;
        for(char c: s1){
            freq[c]++;
        }
        for(int i=0;i<s2.size();i++){
            freq[s2[i]]--;
            if(i>=s1.size()){
                freq[s2[i-s1.size()]]++;
                if(freq[s2[i-s1.size()]]==0){
                    freq.erase(s2[i-s1.size()]);
                }
            }
            if(freq[s2[i]]==0){
                freq.erase(s2[i]);
            }
            
            if(freq.size()==0){
                return true;
            }
        }
        
        return false;
    }
};