[LeetCode] Substring with Concatenation of All Words 串联所有单词的子串

 

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).

 

这道题让我们求串联所有单词的子串，就是说给定一个长字符串，再给定几个长度相同的单词，让我们找出串联给定所有单词的子串的起始位置，还是蛮有难度的一道题。这道题我们需要用到两个哈希表，第一个哈希表先把所有的单词存进去，然后从开头开始一个个遍历，停止条件为当剩余字符个数小于单词集里所有字符的长度。这时候我们需要定义第二个哈希表，然后每次找出给定单词长度的子串，看其是否在第一个哈希表里，如果没有，则break，如果有，则加入第二个哈希表，但相同的词只能出现一次，如果多了，也break。如果正好匹配完给定单词集里所有的单词，则把i存入结果中，具体参见代码如下：

 

class Solution {
public:
    vector<int> findSubstring(string s, vector<string>& words) {
        vector<int> res;
        if (s.empty() || words.empty()) return res;
        int n = words.size(), m = words[0].size();
        unordered_map<string, int> m1;
        for (auto &a : words) ++m1[a];
        for (int i = 0; i <= (int)s.size() - n * m; ++i) {
            unordered_map<string, int> m2;
            int j = 0; 
            for (j = 0; j < n; ++j) {
                string t = s.substr(i + j * m, m);
                if (m1.find(t) == m1.end()) break;
                ++m2[t];
                if (m2[t] > m1[t]) break;
            }
            if (j == n) res.push_back(i);
        }
        return res;
    }
};

 

 

 参考资料：

http://yucoding.blogspot.com/2013/09/leetcode-question-106-substring-with.html

http://blog.unieagle.net/2012/10/28/leetcode%E9%A2%98%E7%9B%AE%EF%BC%9Asubstring-with-concatenation-of-all-words/