Leetcode(30) Substring with Concatenation of All Words

发表于 | 更新于 | 分类于

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.

Example 1:

Input: s = “barfoothefoobarman”, words = [“foo”,”bar”] Output: [0,9]

Example 2:

Input: s = “wordgoodstudentgoodword”, words = [“word”,”student”] Output: []

解法：

题意有点绕，意思是是给你一个字符串，和一个字符串的数组，需要返回一个该字符串的索引组成的数组，其中字符串数组中各成员长度要求一致，返回的索引有如下性质：

从每个索引开始，长度为L的字串需要精确包含字符串数组中的所有字符串（不多不少）。L 为字符串数组中所有字符串长度之和。

思路是：先用一个Map结构，记录字符串数组中各个字符串出现的个数，用于之后判断子串中各字符串是否出现够数，之后，从头开始遍历字符串，找和字符串数组中的字符串相同的字串，找到后map中的值减一，否则重新初始化map，从下一个字符开始遍历。如果map中所有的值都为0，则找到了一个符合条件的子串，索引压入数组。

代码如下：

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
class Solution {
    public List<Integer> findSubstring(String s, String[] words) {
        Map<String, Integer> wordcount = new HashMap<String, Integer>();
        List<Integer> result = new ArrayList<Integer>();
        if(s == "" || words.length == 0){
            return result;
        }
        int wordlen = words[0].length();
        int word_count = words.length;
        for (String word : words) {
            if(word.length()!=wordlen){
                return result;
            }
            if (!wordcount.containsKey(word)) {
                wordcount.put(word, 1);
            } else {
                wordcount.put(word, wordcount.get(word) + 1);
            }
        }

        for (int i = 0; i <= s.length() - wordlen; i++) {
            String tmpsub = s.substring(i, i + wordlen);
            Map<String, Integer> tmpmap = new HashMap<>();
            tmpmap.putAll(wordcount);
            if (!tmpmap.containsKey(tmpsub)) {
                continue;
            } else {
                int remain_word_count = word_count;
                int j = i;
                while (remain_word_count > 0) {
                    if (tmpmap.get(tmpsub) > 0) {
                        tmpmap.put(tmpsub, tmpmap.get(tmpsub) - 1);
                        j += wordlen;
                        remain_word_count -= 1;
                        if(j>s.length() - wordlen){
                            break;
                        }
                        tmpsub = s.substring(j, j + wordlen);
                        if (!tmpmap.containsKey(tmpsub)){
                            break;
                        }
                    } else {
                        break;
                    }
                }
                if(remain_word_count == 0){
                    result.add(i);
                }

            }
        }
        return result;
    }
}