Leetcode||30.Substring with Concatenation of All Words

30. Substring with Concatenation of All Words

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• Difficulty: Hard
• Contributors: Admin

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) ins 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).

讲道理，这题目，真的是……题意好理解，找字符串中的放一起的子串，标记位置。超时了好多次，声明一下string s 改为S，words改为L

代码如下：

public class Solution {    public List<Integer> findSubstring(String S, String[] L) {        List<Integer> ret = new ArrayList<Integer>();        int slen = S.length(), llen = L.length;        if (slen <= 0 || llen <= 0)            return ret;        int wlen = L[0].length();        HashMap<String, Integer> words = new HashMap<String, Integer>();        for (String str : L) {            if (words.containsKey(str)) {                words.put(str, words.get(str) + 1);            } else {                words.put(str, 1);            }        }        for (int i = 0; i < wlen; ++i) {            int left = i, count = 0;            HashMap<String, Integer> tmap = new HashMap<String, Integer>();            for (int j = i; j <= slen - wlen; j += wlen) {                String str = S.substring(j, j + wlen);                if (words.containsKey(str)) {                    if (tmap.containsKey(str)) {                        tmap.put(str, tmap.get(str) + 1);                    } else {                        tmap.put(str, 1);                    }                    if (tmap.get(str) <= words.get(str)) {                        count++;                    } else {                        while (tmap.get(str) > words.get(str)) {                            String tmps = S.substring(left, left + wlen);                            tmap.put(tmps, tmap.get(tmps) - 1);                            if (tmap.get(tmps) < words.get(tmps)) {                                count--;                            }                            left += wlen;                        }                    }                    if (count == llen) {                        ret.add(left);                        String tmps = S.substring(left, left + wlen);                        tmap.put(tmps, tmap.get(tmps) - 1);                        count--;                        left += wlen;                    }                } else {                    tmap.clear();                    count = 0;                    left = j + wlen;                }            }        }        return ret;}