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.

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

You should return the indices: [0,9].

分析：

第一种方式：把words加到hashmap里，然后从i开始，逐一查看是否每个word都在hashmap里，如果存在，则把hashmap中word所对应的count-1，如果hashmap为空，这表示所有的words都已经获取到了，把starting Index加入到list。

但是这种方式过不了。擦。

public class Solution {
    public List<Integer> findSubstring(String s, String[] words) {
        List<Integer> list = new ArrayList<>();
        if (s == null || s.length() == 0 || words == null || words.length == 0)
            return list;

        int length = words[0].length();
        for (int i = 0; i < s.length(); i++) {
            Map<String, Integer> map = getMap(words);
            int index = i;
            boolean allContained = true;
            while (!map.isEmpty()) {
                if (index + length > s.length()) {
                    allContained = false;
                    break;
                }
                String sub = s.substring(index, index + length);
                if (map.containsKey(sub)) {
                    map.put(sub, map.get(sub) - 1);
                    if (map.get(sub) == 0) {
                        map.remove(sub);
                    }
                    index += length;
                } else {
                    allContained = false;
                    break;
                }
            }

            if (allContained) {
                list.add(i);
            }
        }
        return list;
    }

    private Map<String, Integer> getMap(String[] words) {
        Map<String, Integer> map = new HashMap<>();

        for (String str : words) {
            map.put(str, map.getOrDefault(str, 0) + 1);
        }
        return map;
    }
}

还有一种方法，就是先一个window去覆盖S，使得被words里的word都在被覆盖的substring里，那么对于下一个，我们需要把window左边的部分右移一个word出来，这样右边可以继续移动。如果我们当前右边要加入的word已经加够了，或者不存在于words里，我们也需要移动window左边的pointer。很可惜的是，这种方法也过不了，不管了。

public List<Integer> findSubstring(String s, String[] words) {
        Map<String, Integer> all = getMap(words);
        List<Integer> list = new ArrayList<>();
        if (s == null || s.length() == 0 || words == null || words.length == 0)
            return list;

        int length = words[0].length();
        for (int i = 0; i < length; i++) {
            Map<String, Integer> map = getMap(words);
            int index = i;
            int starting = i;
            while (index + length <= s.length()) {
                String sub = s.substring(index, index + length);
                if (map.containsKey(sub)) {
                    map.put(sub, map.get(sub) - 1);
                    if (map.get(sub) == 0) {
                        map.remove(sub);
                    }
                    // find all the words in the hashmap
                    if (map.isEmpty()) {
                        list.add(starting);
                        map.put(s.substring(starting, starting + length), 1);
                        starting = starting + length;
                    }
                } else {
                    if (all.containsKey(sub)) {
                        map.put(s.substring(starting, starting + length),
                                map.getOrDefault(s.substring(starting, starting + length), 0) + 1);
                        starting = starting + length;
                        continue;
                    } else {
                        map = getMap(words);
                        starting = index + length;
                    }
                }
                index += length;
            }
        }
        return list;
    }

    private Map<String, Integer> getMap(String[] words) {
        Map<String, Integer> map = new HashMap<>();

        for (String str : words) {
            map.put(str, map.getOrDefault(str, 0) + 1);
        }
        return map;
    }