Substring with Concatenation of All Words

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嗯先贴一个最初的实现，思路是对的，但是编码能力太差，写出来有问题。

class Solution {public:    vector<int> findSubstring(string S, vector<string> &L) {        int llen = L[0].length();        vector<int> res;        if (S == "")            return res;        vector<bool> used;        for (int i = 0; i < L.size(); ++i)            used.push_back(false);        for (int i = 0; i < S.length(); ++i) {            for (int j = 0; j < L.size(); ++j)                used[j] = false;            int j;            for (j = 0; j < L.size() && i + (j + 1) * llen <= S.length(); ++j) {                if (!isValid(S.substr(i + j * llen, llen), L, used)) {                    i = i + j * llen;                    break;                }            }            if (j == L.size())                res.push_back(i);        }    }    bool isValid(string s, vector<string> &L, vector<bool> &used) {        int index = -1;        for (int i = 0; i < L.size(); ++i) {            if (s == L[i] && used[i] == false)                index = i;        }        if (index == -1)            return false;        used[index] = true;        return true;    }};

目前的结论是，我之前的思路是正确的，使用map可以大大加快速度。并且，由于STL效率差，因此能不重复做的工作坚决不重复做。

class Solution {public:vector<int> findSubstring(string S, vector<string> &L) {vector<int> res;if (S == "")return res;map<string, int> words;map<string, int> curStr;for (int i = 0; i < L.size(); ++i)++words[L[i]];int len = L.size() * L[0].length();for (int i = 0; i + len <= S.length(); ++i) {if (isValid(S.substr(i, len), L, words, curStr))res.push_back(i);}return res;}bool isValid(string s, vector<string> &L, map<string, int> &words, map<string, int> &curStr) {curStr.clear();int wordLen = L[0].length();for (int i = 0; i + wordLen <= s.length(); i += wordLen) {string sub = s.substr(i, wordLen);if (words.find(sub) == words.end())return false;++curStr[sub];if (curStr[sub] > words[sub])return false;}return true;}};

http://oj.leetcode.com/problems/substring-with-concatenation-of-all-words/

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