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.
Example 1:
Input:
s = "barfoothefoobarman",
words = ["foo","bar"]
Output: [0,9]
Explanation: Substrings starting at index 0 and 9 are "barfoor" and "foobar" respectively.
The output order does not matter, returning [9,0] is fine too.
Example 2:
Input:
s = "wordgoodgoodgoodbestword",
words = ["word","good","best","word"]
Output: []
特殊收获：
1.substr函数
返回一个新建的初始化为string对象的子串的拷贝string对象。
子串是，在字符位置pos开始，跨越len个字符（或直到字符串的结尾，以先到者为准）对象的部分。
2.unordered_map哈希表 count函数
Count函数
size_type count ( const key_type& key ) const
count函数用以统计key值在unordered_map中出现的次数。实际上，c++ unordered_map不允许有重复的key。因此，如果key存在，则count返回1，如果不存在，则count返回0.
3.引入一个新结构类型时，应该想需要什么头文件外部依赖，比如hash表，this structure need extra support #include<unordered_map>
4.How do you actually understand a function
eg:
if(!struc.count(t)) break;
In this function,first I thought need add sth after struc(an unordered_map),but actually this function is to judge whether t in the hashmap,use count function so t need to be as a variable pass into function count.Know where should place function is the first step
5.reset(清零) is an important habit.Especially in counting,many problem comes because of not reset.
6.Assignment is the most effective weapon for supporting ideas
7.Think special case by yourself rather than rely on System judgement
In this code i++ first I use i+=n*len,i+=len,but it has a special case is such as "aaaaaaa" '"aa"this kinds of duplicate characterised
so at last we can only use i++
#include <iostream> #include<vector> #include<iostream> #include<string> #include<stdio.h> #include<string.h> #include<iomanip> #include<vector> #include<list> #include<queue> #include<algorithm> #include<stack> #include<map> #include<unordered_map> using namespace std; class Solution { public: vector<int> change(string s,vector<string>& word) { vector<int> res; int j=0; if(s.size()==0||word.size()==0) return res; int n=word.size(),len=word[0].size(); unordered_map<string,int> struc; //build a hashMap to store structure for(auto wor:word) { ++struc[wor]; } //prove insert correct for(auto wor:word) { cout<<"struc: "<<wor<<" value: "<<struc[wor]<<endl; //++struc[wor]; } for(int i=0;i<=(int)s.size()-n*len;) { string sent=s.substr(i,n*len); cout<<"sent : "<<sent<<endl; unordered_map<string,int> test; for(j=0;j<n*len;j+=len) { string t=sent.substr(j,len); cout<<"t : "<<t<<endl; //if exist count==1,add !==0 break if not exist break; if(!struc.count(t)) break; ++test[t]; if(test[t]>struc[t]) break; } if(j==n*len) res.push_back(i); } return res; } }; int main() { string s1; cin>>s1; vector<string> word; word.push_back("bar"); word.push_back("foo"); word.push_back("the"); //word.push_back("word"); //word.push_back("ssd"); vector<int> res; Solution s; res=s.change(s1,word); for(auto re:res) cout<<re<<endl; return 0; }
This method used to first cut sent from s,then cut t from sent.Add extra space.Lower the speed ,if we let j=0;j<n;j++,in this way,we only need cut and space once.Largely save and space faster a little.
The only thing I want to consist in coding is repetition,First clear your mind,In a word,close the book and don't open it until you finish,compare the performance difference between you and it.Finally,make progress!
Improve the code as following:
class Solution { public: vector<int> findSubstring(string s,vector<string>& word) { vector<int> res; int j=0; if(s.empty()||word.empty()) return {}; int n=word.size(),len=word[0].size(); unordered_map<string,int> struc; //build a hashMap to store structure for(auto &wor:word) { ++struc[wor]; } for(int i=0;i<=(int)s.size()-n*len;++i) { //string sent=s.substr(i,n*len); unordered_map<string,int> test; for(j=0;j<n;++j) { string t=s.substr(i+j*len,len); if(!struc.count(t)) break; ++test[t]; if(test[t]>struc[t]) break; } if(j==n) res.push_back(i); } return res; } };
