438. Find All Anagrams in a String

问题描述：

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".

 

解题思路：

仔细读题，是让我们找，在s中p的变位词。

我一开始以为让找s内存在的所有的变位次，当时有点懵逼

这样的话可以用滑动窗口来做，移动r：

　　1.当s[r] 的字符不属于 p的时候，要将l移动到r的位置，注意改变cnt

　　2.当s[r]的字符属于p的时候，需要检查，当前target[s[r]]的值是否大于0，若等于0，则需要移动l到大于0为止，同样不要忘记改变cnt

　　3.检查cnt是否为0，若为n，则找到变位词。

 

代码：

class Solution {
public:
    vector<int> findAnagrams(string s, string p) {
        vector<int> ret;
        if(s.empty())
            return ret;
        unordered_map<char, int> target;
        for(char c:p){
            target[c]++;
        }
        int l = 0;
        int r = 0;
        int cnt = p.size();
        int sLen = s.size();
        while(r < sLen){
            if(target.count(s[r]) == 0){
                while(l < r){
                    target[s[l++]]++;
                    cnt++;
                }
                r++;
                l = r;
            }else{
                if(target[s[r]] > 0){
                    target[s[r++]]--;
                    cnt--;
                }else{
                    while(l < r && target[s[r]] == 0) {
                        target[s[l++]]++;
                        cnt++;
                    }
                }
                if(cnt == 0){
                    ret.push_back(l);
                    target[s[l++]]++;
                    cnt++;
                }
            }
        }
        return ret;
    }
};