leetcode_c++：哈希：Substring with Concatenation of All Words(030)

题目

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).

给定字符S和单词表，单词长度都一样，找出所有s的字串，字串由所有的单词组成

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算法

O（n*m）

• L里每个单词长度一样，写循环就方便了很多。先初始化一个map，统计L里每个单词出现的次数。每次循环如果某个单词没出现或者超出L中出现的错误就中断。

• vector的写法

• http://blog.csdn.net/acs713/article/details/44939815

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class Solution {public:    vector<int> findSubstring(string S, vector<string> &L) {        int l_size = L.size();        if (l_size <= 0) {            return vector<int>();        }        vector<int> result;        map<string, int> word_count;        int word_size = L[0].size();        int i, j;        for (i = 0; i < l_size; ++i) {            ++word_count[L[i]];        }        map<string, int> counting;        for (i = 0; i <= (int)S.length() - (l_size * word_size); ++i) {            counting.clear();            for (j = 0; j < l_size; ++j) {                string word = S.substr(i + j * word_size, word_size);                if (word_count.find(word) != word_count.end()) {                    ++counting[word];                    if (counting[word] > word_count[word]) {                        break;                    }                }                else {                    break;                }            }            if (j == l_size) {                result.push_back(i);            }        }        return result;    }};