30. Substring with Concatenation of All Words（js）

30. Substring with Concatenation of All Words

You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.

Example 1:

Input:
  s = "barfoothefoobarman",
  words = ["foo","bar"]
Output: [0,9]
Explanation: Substrings starting at index 0 and 9 are "barfoor" and "foobar" respectively.
The output order does not matter, returning [9,0] is fine too.

Example 2:

Input:
  s = "wordgoodgoodgoodbestword",
  words = ["word","good","best","word"]
Output: []
题意：在字符串s中提取字串，当words数组所有项顺序不定的组合成的字符串与字串相等时，将字串的起始索引存入结果返回
代码如下：

/**
 * @param {string} s
 * @param {string[]} words
 * @return {number[]}
 */
var findSubstring = function(s, words) {
    let res=[];
    if(s.length==0 || words.length==0) return res;
    let n=words.length,m=words[0].length;
    let m1={};
//     将所有单词存入哈希表
    for(let w of words){  
        m1[w]?m1[w]++:(m1[w]=1)
    }

    for(let i=0;i<=s.length-n*m;i++){
        let m2={};
        let j=0;
        for(j=0;j<n;j++){
//             获取长度为m的字串t
            let t=s.substr(i+j*m,m);
//             如果m1中不存在则跳出循环
            if(!m1[t]) break;
//             如果m1中存在字串t,则存入m2
            m2[t]?m2[t]++:(m2[t]=1);
//             如果m2中出现的单词次数比m1多，跳出循环
            if(m2[t]>m1[t]) break;
        }
        if(j==n) res.push(i);
    }
    return res;
};