LeetCode第30题：Substring with Concatenation of All Words（C++图文详解）_leetcode 30

30. Substring with Concatenation of All Words
    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: []

题目详解：
1.此题得难度在于words中单词可重复；
2.单纯遍历给定字符串，无法判断；
3.思路是将words中每个单词存入map中，因可重复，故存入unordered_map；
4.因words中单词定长，可算出长度，故遍历给定字符串，只需到（字符串长度减去words中单词总长度）；

代码：

#include "pch.h"
#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;
vector<int> findSubstring(string s, vector<string>& words) {
	if (s.empty() || words.empty()) return {};
	vector<int> resVec;//返回数组
	int n = words.size(), len = words[0].size();
	unordered_map<string, int> wordsMap;
	for (auto &word : words) ++wordsMap[word];
	for (int i = 0; i <= (int)s.size() - n * len; i++) {
		unordered_map<string, int> sMap;
		int j = 0;
		for (j = 0; j < n; j++) {
			string t = s.substr(i + j * len, len);//截取定长子字符串（words中单词定长）
			if (!wordsMap.count(t)) break;//当前截取子字符串不在map中，终止
			++sMap[t];
			if (sMap[t] > wordsMap[t]) break;//子字符串值不相等，终止
		}
		if (j == n) resVec.push_back(i);//连续给定子字符串都相等，则为所要值
	}
	return resVec;
}
int main()
{
	/*string s = "wordgoodgoodgoodbestword";
	vector<string> words = { "word", "good", "best", "word" };*/
	string s = "barfoothefoobarman";
	vector<string> words = {"foo", "bar"};
	vector<int> res = findSubstring(s, words);
	for (int i = 0; i < res.size(); i++)
	{
		cout << res[i];
		cout << "\n";
	}
	
    std::cout << "Hello World!\n"; 
	system("PAUSE");
}
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图文详解：

性能：

总结：此题目主要考察对Map得使用，采用其他方法可能比较费劲；容器得原理和理解，在数据结构中尤为重要；继续加油！