LeetCode Substring with Concatenation of All Words

最新推荐文章于 2019-12-26 08:25:23 发布

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本文介绍了一种高效的字符串匹配算法，该算法能在给定的字符串S中找到由单词列表L构成的所有子串的起始索引，其中L中的所有单词长度相同且每个单词仅使用一次。通过构建单词表并遍历字符串S，实现对符合条件的子串的有效查找。

题目：

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) {
		if (L.size() == 0 || S.size() < L[0].size()* L.size())
			return vector<int>(0);
		vector<int> res;
		unordered_map<string, int> table;
		unordered_map<string, int> cur;
		//建立单词表
		for (int i = 0; i < L.size(); i++) {
			if (table.find(L[i]) == table.end())
				table[L[i]] = 1;
			else
				table[L[i]]++;
		}
		//每个单词长度
		int wl = L[0].size();
		//单词个数
		int wn = L.size();
		for (int i = 0; i <= S.size() - wn*wl; i++) {
			cur.clear();
			int j;
			for (j = 0; j < wn; j++) {
				string s = S.substr(i + j* wl, wl);
				//单词不在表中
				if (table.find(s) == table.end())
					break;
				cur[s]++;
				if (cur[s] > table[s])
					break;
			}
			if (j == wn)
				res.push_back(i);
		}
		return res;
	}
};