LeetCode 30. Substring with Concatenation of All Words

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问题

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

解法

经过测试，words中的字符串存在重复的情况。
因此第一步是统计每种字符串个数。
然后在s中进行枚举，比较当前串是否符合。
在枚举s时使用滑动窗口算法。

typedef unordered_map<string, int>::iterator it;class Solution {public:    vector<int> findSubstring(string s, vector<string>& words) {        vector<int> ret;        if (words.size() < 1)            return ret;        int wdLen = words[0].size();        if (s.size() <wdLen* words.size())            return ret;        unordered_map<string, int> wdCnt;        for(int i=0; i< words.size(); ++i)            wdCnt[words[i]]++;        for (int i=0; i<wdLen; ++i)        {            int left = i;            unordered_map<string, int> found;            int foundNum = 0;            for(int j = i; j + wdLen<=s.size(); j+= wdLen)            {                string sub = s.substr(j, wdLen);                if(wdCnt.find(sub) == wdCnt.end())                {                    found.clear();                    foundNum =0;                    left = j+wdLen;                }else{                    found[sub]++;                    foundNum++;                    if (found[sub] > wdCnt[sub])                    {                        while(found[sub] > wdCnt[sub])                        {                            string ss = s.substr(left, wdLen);                            found[ss]--;                            foundNum--;                            left+= wdLen;                        }                    }                    if (foundNum == words.size())                    {                        ret.push_back(left);                        string ss = s.substr(left, wdLen);                        found[ss]--;                        foundNum--;                        left+=wdLen;                    }                }            }        }        return ret;    }};

实现细节

unordered_map<> clear()

Clear content
All the elements in the unordered_map container are dropped: their destructors are called, and they are removed from the container, leaving it with a size of 0.

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