LeetCode-030 Substring with Concatenation of All Words

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

Example 1:

Input:
s = “barfoothefoobarman”,
words = [“foo”,“bar”]
Output: [0,9]
Explanation: Substrings starting at index 0 and 9 are “barfoor” and “foobar” respectively.
The output order does not matter, returning [9,0] is fine too.

Example 2:

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

Analyse

題目的意思是給定一個字符串s以及一個字符串列表words，讓你找出在s中用words裏面所有單詞排列而成的子串的首位置。
我採用的方法很簡單，就是枚舉s中的所有位置，然後生成一個words列表裏面所有單詞組成的隊列，取出一個，與s匹配，匹配不上的話，就重新壓進隊列，匹配得上的話，位移之後繼續匹配，直到隊列爲空或者隊列裏的元素無減少。稍微做了點剪枝，設s的長度爲s_len，words裏面的單詞長度總和爲word_len，那麼對s的枚舉只需要到s_len-word_len就夠了。時間複雜度爲O(s_len * word_len * word_len)，效率是比較差的，然後測試的時候也是剛好在時間點上飄過。優化方法有對單詞進行哈希處理，這樣可以減少比較運算。不過嫌麻煩就沒做了。

Code

class Solution {
public:
    bool isCorrect(string& s, string& word, int& temp_pos) {
        int word_pos = 0;
        while (word_pos < word.length()) {
            if (s[temp_pos + word_pos] == word[word_pos]) {
                word_pos++;
            } else {
                return false;
            }
        }
        return true;
    }
    vector<int> findSubstring(string s, vector<string>& words) {
        int len = words.size();
        vector<int> ans;
        if (s.length() == 0 || words.size() == 0) return ans;
        int s_pos = 0;
        int total_len = 0;
        queue<string> words_queue;
        for (int i = 0; i < len; i++) {
            words_queue.push(words[i]);
            total_len += words[i].length();
        }
        while (s_pos + total_len <= s.length()) {
            queue<string> words_temp = words_queue;
            int temp_pos = s_pos;
            int count = 0, pre_len = len;
            while (!words_temp.empty() && count != pre_len) {
                string word = words_temp.front();
                words_temp.pop();
                if (isCorrect(s, word,temp_pos)) {
                    temp_pos += word.length();
                    pre_len--;
                    count = 0;
                } else {
                    words_temp.push(word);
                    count++;
                }
            }
            if (words_temp.empty()) {
                ans.push_back(s_pos);
            }
            s_pos++;
        }
        return ans;
    }
};