1065. Index Pairs of a String 字符串的索引对

• 作者： 负雪明烛
• id： fuxuemingzhu
• 个人博客：https://fuxuemingzhu.cn/

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题目地址：https://leetcode-cn.com/problems/index-pairs-of-a-string/

题目描述

Given a text string and words (a list of strings), return all index pairs [i, j] so that the substring text[i]...text[j] is in the list of words.

Example 1:

Input: text = "thestoryofleetcodeandme", words = ["story","fleet","leetcode"]
Output: [[3,7],[9,13],[10,17]]

Example 2:

Input: text = "ababa", words = ["aba","ab"]
Output: [[0,1],[0,2],[2,3],[2,4]]
Explanation:
Notice that matches can overlap, see "aba" is found in [0,2] and [2,4].

Note:

1. All strings contains only lowercase English letters.
2. It’s guaranteed that all strings in words are different.
3. 1 <= text.length <= 100
4. 1 <= words.length <= 20
5. 1 <= words[i].length <= 50 Return the pairs [i,j] in sorted order (i.e. sort them by their first coordinate in case of ties sort them by their second coordinate).

题目大意

给出 字符串 text 和 字符串列表 words, 返回所有的索引对 [i, j] 使得在索引对范围内的子字符串 text[i]...text[j]（包括 i 和 j）属于字符串列表 words。

解题方法

遍历

暴力遍历所有的字符串子串，看其是否在words中。为了加速查找效率，使用的set。

C++代码如下：

class Solution {
public:
    vector<vector<int>> indexPairs(string text, vector<string>& words) {
        unordered_set<string> wordset(words.begin(), words.end());
        const int N = text.size();
        vector<vector<int>> res;
        for (int i = 0; i < N; ++i) {
            for (int j = i; j < N; ++j) {
                string cur = text.substr(i, j - i + 1);
                if (wordset.count(cur)) {
                    res.push_back({i, j});
                }
            }
        }
        return res;
    }
};

日期

2019 年 9 月 18 日 —— 今日又是九一八