LeetCode 30. Substring with Concatenation of All Words

原题链接在这里: https://leetcode.com/problems/substring-with-concatenation-of-all-words/description/

题目： 

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).

题解：

类似Longest Substring Without Repeating Characters, 这类快慢指针维护substring的方法在Minimum Window Substring里有总结.

需要先生成global map计算words中每个word 的frequency. 快慢指针围成的window里也有个小的map计算map中出现在words的word, 当出现一个就count++.

当count是words数目和时就找到了一个结果.

检查当前词是否是words里的词时外层循环是多种取词方法，以例子来说就是分成：

|bar|foo|the|foo|bar|man

b|arf|oot|hef|oob|arm|an

ba|rfo|oth|efo|oba|rma|n

Note: s.substring(startIndex, endIndex) 其中的endIndex必须比s string本身长度小.

Time Complexity: O(s.length()).

Space: O(words.length). 

AC Java:

class Solution {
    public List<Integer> findSubstring(String s, String[] words) {
        List<Integer> res = new ArrayList<Integer>();
        if(words == null || words.length == 0 || s == null || s.length() == 0){
            return res;
        }
        
        // 计算原有word frequency
        HashMap<String, Integer> hm = new HashMap<String, Integer>();
        for(String word : words){
            hm.put(word, hm.getOrDefault(word, 0)+1);
        }
        
        int wordLen = words[0].length();
        int count = 0;
        int walker = 0;
        
        // 从等长word长度的开始到结尾挨个试，不需要再往后，因为那已经被长度开始里面的runner下一跳包括
        for(int i = 0; i<wordLen; i++){
            walker = i;
            count = 0;
            HashMap<String, Integer> windowHm = new HashMap<String, Integer>();
            int max = s.length()-wordLen+1;
            
            for(int runner = walker; runner < max; runner += wordLen){
                String cur = s.substring(runner, runner+wordLen);
                if(hm.containsKey(cur)){
                    // cur string在words中
                    windowHm.put(cur, windowHm.getOrDefault(cur, 0)+1);
                    if(windowHm.get(cur) <= hm.get(cur)){
                        count++;
                    }else{
                        // cur这个词在window里出现的次数 多余 原来words中的次数
                        while(windowHm.get(cur) > hm.get(cur)){
                            String walkerStr = s.substring(walker, walker+wordLen);
                            windowHm.put(walkerStr, windowHm.get(walkerStr)-1);
                            if(windowHm.get(walkerStr) < hm.get(walkerStr)){
                                count--;
                            }
                            walker += wordLen;
                        }
                    }
                    
                    // 全部覆盖，向前移动walker
                    if(count == words.length){
                        res.add(walker);
                        String walkerStr = s.substring(walker, walker+wordLen);
                        windowHm.put(walkerStr, windowHm.get(walkerStr)-1);
                        count--;
                        walker += wordLen;
                    }
                }else{
                    // cur 不在words中
                    windowHm.clear();
                    count = 0;
                    walker = runner+wordLen;
                }
            }
        }
        
        return res;
    }
}