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

思路1：

 首先s1全排列，每一个全排列 查看是否是s2的子串。果然...超时，复杂度太高。

bool checkInclusion(string s1, string s2)
{
    if(s1.size()>s2.size())return false;
    char* s1char = new char(s1.size()+1);
    s1.copy(s1char,s1.size(),0);
    char* s1char2= new char(s1.size()+1);
    s1.copy(s1char2,s1.size(),0);
    do
    {
        if(s2.find(s1char) != string::npos) return true;
    }while(next_permutation(s1char,s1char+strlen(s1char)));
    do
    {
        if(s2.find(s1char2) != string::npos) return true;
    }while(prev_permutation(s1char2,s1char2+strlen(s1char2)));
return false;
}

思路2：

学习了已经AC的大神的代码，发现他们无一例外都是用hash表格来做的。

思路就是：观察字符串s1和s2，如果s1的全排列是s2的子串，则s1的所有字符均出现在s2中，而且这些字符连续。

于是用一个类似于滑动窗口的区间去统计此区间内的字符，区间的长度取为s1的长度。

如果在s2中存在一个区间，里面的字符与s1的字符全部都一样，则说明s1通过全排列后一定能够得到s2中的子串。也即满足题意。

bool checkInclusion2(string s1, string s2)
{
    int m = s1.length(),n = s2.length();
    if(m>n)return false;
    vector<int>maps1(26),maps2(26);
    for(int i=0;i<m;i++)
    {
        maps1[s1[i] - 'a'] ++;
        maps2[s2[i] - 'a'] ++;
    }
    if(maps1 == maps2) return true; //统计前m个字符是否符合条件
    for(int i = 0;i+m<n;i++)
    {
         maps2[s2[i] - 'a'] --;
         maps2[s2[i+m] - 'a'] ++;
         if(maps1 == maps2) return true;
    }
    return false;
}