leetcode-30.Substring wih Concatenation of All Words

题目

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 = "wordgoodstudentgoodword",
words = ["word","student"]
Output: []

思路

如果需要找的words都是等长的，直接利用滑动窗口以words为单位检索

否则，需要先利用滑动窗口以letter为单位检索符合letter统计的区间，然后再在数量和对象上判断该区间是否为目标words组成

代码

#include "stdafx.h"
#include <list>
#include <vector>
#include <map>
#include <iostream>
using namespace std;


void showmap(map<char, int>& input) {
	map<char, int>::iterator it;
	for (it = input.begin();it!=input.end();it++) {
		cout << it->first<<'\t'<<it->second << endl;
	}
}

vector<int> findSubstring(string S, vector<string> &L) {
	vector<int> ans,ans_result;
	int n = S.size();
	//make statistics of the letter of L
	map<char, int> cmap,curmap;
	for (string s : L) {
		for (char c : s) {
			if (cmap.find(c) != cmap.end()) {
				cmap[c] ++;
			}
			else {
				cmap[c] =1;
			}
		}
	}
	//showmap(cmap)；
	int count = 0;
	int start = 0;
	int M = cmap.size();
	//sliding windows, searching for the fragments that in line with the statistics of letter
	for (int i = 0; i < n; i++) {
		char c = S[i];
		if (cmap.find(c) != cmap.end()) {
			if (curmap.find(c) != curmap.end()) { curmap[c]++; }
			else { curmap[c] = 1; }
			if (curmap[c] == cmap[c]) { count++; }
			while (curmap[c] > cmap[c]) {
				char t = S[start];
				if (curmap[t] == cmap[t]) count--;
				curmap[t]--;
				start++;
			}
			if (count == cmap.size()) {
				ans.push_back(start);
				char t = S[start];
				count--;
				curmap[t]--;
				start++;
			}
		}
		else {
			curmap.clear();
			start = i + 1;
			count = 0;
		}

	}

	//make statistics of the words and the words' length of L
	map<string, int> wmap, wurmap;
	map<int, int> wlmap, wlurmap;
	int wl_total = 0;
	for (string s : L) {
		if (wmap.find(s) != wmap.end()) { wmap[s]++; }
		else {wmap[s] = 1;}
		if (wlmap.find(s.length()) == wlmap.end()) { wlmap[s.length()] ++; }
		else { wlmap[s.length()] = 1; }
		wl_total += s.length();
	}

	for (int i = 0; i < ans.size(); i++) {
		map<int, int>::iterator itn;
		start = ans[i];
		bool flag = false;//whether the word have been verified as a member of wmap
		bool flag_total = true;//whether a segment is a final result 
		int count = 0;
		wurmap.clear();
		//cout << start << '\t' << ans[i] + wl_total<<endl;
		//judge from the start to the end of the fragment
		while ((start<(ans[i] + wl_total)) && flag_total) {
			//for each possible length
			for (itn = wlmap.begin(); itn != wlmap.end(); itn++) {
				//if the word has not been verified as the member of wmap
				if (flag == false) {
					//capture a word 
					string tmp = S.substr(start, itn->first);
					//judge if the word is a member of wmap
					if (wmap.find(tmp) != wmap.end()) {
						//use wurmap to make statistics of the words in the fragment
						if (wurmap.find(tmp) != wurmap.end()) { wurmap[tmp]++; }
						else { wurmap[tmp] = 1; }
						//if the numbers of words equal in wurmap and wmap , add count, meaning that all the same word have been find out 
						if (wurmap[tmp] == wmap[tmp]) { count++; }
						else { if (wurmap[tmp] > wmap[tmp]) { flag_total = false; } }// if the number of some word exceed that in wmap, it means this fragment is not a right final result
						//this word has been verified
						flag = true;
						//move to next word' start position of this fragment
						start += itn->first;
					}
				}
				else { break; }
			}
			//if all possible length are not valid, that means no word in L has been find in the current start position and the verify process is finished, this fragment is a false positive 
			if (flag == false) {flag_total = false; }
			else { flag = false; }
		}
		
		if (start == ans[i] + wl_total&&flag_total == true) { ans_result.push_back(ans[i]); }

	}

	return ans_result;
}

int main()
{
	vector<string> L;
	L.push_back("apple");
	L.push_back("bee");
	cout << L.size() << endl << L[0].size();
	vector<int> result = findSubstring("pineapplebeebirdsquarebeeapple", L);
	cout << endl << result.size() << endl;
    return 0;
}

问题

以上方法并不能解决以下特殊情况：

Example 3:

Input:
s = "goodeffecteffectivemanyappleeffectivegoodeffect",
words = ["effect","effective","good"]
Output: [0,28]

因为effect 和effective 长度不同，但是前面部分一样，这样的话以以上的方法无法解决。所以这个问题还是目标words等长才可能在短时间内把问题解决，不然情况太复杂了