Description

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.

For example, given:
s: "barfoothefoobarman"
words: ["foo", "bar"]

You should return the indices: [0,9].
(order does not matter).

Solution

Slide-window + HashMap, time O(n * m), space O(1)

首先将words缓存在HashMap中，然后遍历s，把每个i作为startIndex开始，往后面尽量扩展即可。

class Solution {
    public List<Integer> findSubstring(String s, String[] words) {
        List<Integer> startIdx = new ArrayList<>();
        int targetLen = 0;
        Map<String, Integer> map = new HashMap<>();
        for (String word : words) {
            targetLen += word.length();
            map.put(word, map.getOrDefault(word, 0) + 1);
        }
        
        if (s.length() < targetLen || words.length == 0) {
            return startIdx;
        }
        
        int wordLen = words[0].length();
        
        for (int i = 0; i <= s.length() - targetLen; ++i) {
            Map<String, Integer> need = new HashMap<>(map);
            boolean isValid = true;
            
            for (int j = i; j < i + targetLen; j += wordLen) {
                String curr = s.substring(j, j + wordLen);
                if (need.getOrDefault(curr, 0) == 0) {
                    isValid = false;
                    break;
                }
                need.put(curr, need.get(curr) - 1);
            }
            
            if (isValid) {
                startIdx.add(i);
            }            
        }
        
        return startIdx;
    }
}