**LeetCode 30. Substring with Concatenation of All Words

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

滑动窗口法。解析看这里：

http://www.2cto.com/kf/201406/311648.html

算法时间复杂度为O（n*k)）n是字符串的长度，k是单词的长度

#include <cstdio>#include <iostream>#include <vector>#include <algorithm>#include <map>using namespace std;class Solution {public:    vector<int> findSubstring(string s, vector<string>& words) {        map<string,int> wd;        for(int i=0;i<words.size();i++)            wd[ words[i] ] ++;        vector <int> ret;        if( words.size() == 0 ||           words[0].size() > s.size() )return ret;        int sz = words[0].size();        for(int i=0;i<sz;i++) {            int winSt = i, cnt=0;            map <string, int> found;            for(int winEnd = i; winEnd <= s.size()-sz; winEnd += sz ) {                //s.size()-sz because word = s.substr(winEnd, sz);                //也就是最后一个位置winEnd开始放words[0]                string word = s.substr(winEnd, sz);                if(wd.find(word) != wd.end()) {                    if(wd[word] > found[word]) {                        found[word] ++;                        cnt ++;                    } else {                        string tmp;                        while( (tmp=s.substr(winSt, sz)) != word ) {                            winSt += sz;                            found[tmp] --;                            cnt --;                        }                        winSt += sz;                        //here can't because -- and insert word , so you needn't do anything except winSt+=sz                        //found[tmp] --;                        //cnt --;                    }                    if(cnt == words.size()) {  //从下一个位置开始                        ret.push_back(winSt);                        found[ s.substr(winSt, sz) ] --;                        winSt += sz;                        cnt --;                    }                } else {                    found.clear();                    winSt = winEnd + sz;                    cnt=0;                }            }        }        return ret;    }};int main() {    freopen("l30.txt", "r", stdin);    int n;    string s,in;    while(cin >> n) {        vector<string>svec;        for(int i=0;i<n;i++) {            cin >> in;            svec.push_back(in);        }        cin >> s;        Solution so;        vector<int>ans = so.findSubstring(s, svec);        for(int i=0;i<ans.size();i++) {            cout << ans[i] << ", ";        }        cout << "ans end" << endl;        cout << endl;    }    return 0;}

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