leetcode substring-with-concatenation-of-all-words-优快云博客

本文链接：https://blog.youkuaiyun.com/pynash123/article/details/84638627

本文介绍了一种用于寻找字符串中特定子串的方法，特别是当输入字符串包含由相同长度单词组成的列表时，该算法能找出所有可能的起始索引，使得这些索引对应的子串恰好是由给定单词列表中的单词按任意顺序拼接而成。

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You are given a string, S, and a list of words, L, that are all of the same length. Find all starting indices of substring(s) in S that is a concatenation of each word in L exactly once and without any intervening characters.
For example, given:
S:“barfoothefoobarman”
L:[“foo”, “bar”]
You should return the indices:[0,9].
(order does not matter).

class Solution {
public:
    vector<int> findSubstring(string S, vector<string> &L) 
    {
        vector<int> res;
        if (S.size() == 0 || L.size() == 0) return res;
        int size = L.size()*L[0].size();
        if (S.size()<size) return res;
        for (int i=0;i<=S.size()-size;i++)
            if (isone(S.substr(i), L))
                res.push_back(i);
        return res;
    }
    bool isone(string S, vector<string> L)
    {
        if (L.size() == 0) return true;
        int size = L[0].size();
        vector<string>::iterator iter;
        for(int i=0;i<L.size();i++)
        {
            iter = find(L.begin(),L.end(),S.substr(0, size));
            if (iter == L.end())
                return false;
            else
            {
                L.erase(iter);
                return isone(S.substr(size), L);
            }
        }
        return false;//这里返回时因为网页编译报错
    }
};

测试

#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
using namespace std;
class Solution {
public:
    vector<int> findSubstring(string S, vector<string> &L) 
    {
        vector<int> res;
        if (S.size() == 0 || L.size() == 0)
            return res;
        int size = L.size()*L[0].size();
        if (S.size()<size)
            return res;
        for (int i=0;i<=S.size()-size;i++)
        {
            if (isone(S.substr(i), L))
                res.push_back(i);
        }
        return res;
    }
    
    bool isone(string S, vector<string> L)
    {
        
        if (L.size() == 0)
            return true;
        int size = L[0].size();
        string tmp = S.substr(0, size);
        vector<string>::iterator iter;
        for(int i=0;i<L.size();i++)
        {
            iter = find(L.begin(),L.end(),tmp);
            if (iter == L.end())
                return false;
            else
            {
                L.erase(iter);
                return isone(S.substr(size), L);
            }
        }
        return false;
    }
};

int main()
{
    Solution s;
    string S = "barfoothefoobarman";
    vector<string> vec;
    vec.push_back("foo");
    vec.push_back("bar");
    vector<int> res = s.findSubstring(S, vec);
    for(int i=0;i<res.size();i++)
        cout<<res[i]<<endl;
    cout<<endl;
    return 0;
}