【LeetCode】245. Shortest Word Distance III 解题报告 (C++)

目录

题目地址：https://leetcode-cn.com/problems/shortest-word-distance-iii/

题目描述

Given a list of words and two words word1 and word2, return the shortest distance between these two words in the list.

word1 and word2 may be the same and they represent two individual words in the list.

Example:

Assume that words = ["practice", "makes", "perfect", "coding", "makes"].

Input: word1 = “makes”, word2 = “coding”
Output: 1
Input: word1 = "makes", word2 = "makes"
Output: 3

Note:

  • You may assume word1 and word2 are both in the list.

题目大意

给定一个单词列表和两个单词 word1 和 word2，返回列表中这两个单词之间的最短距离。

word1 和 word2 是有可能相同的，并且它们将分别表示为列表中两个独立的单词。

解题方法

字典+暴力检索

字典保存每个单词的下标，然后分类讨论：

  1. 如果两个单词不同，那么直接暴力检索这两个的所有坐标的差值绝对值，找最小的。
  2. 如果两个单词相同，那么相当于从一个有序数组中找到相邻数字的最小差值，需要一次遍历。

C++代码如下：

class Solution {
public:
    int shortestWordDistance(vector<string>& words, string word1, string word2) {
        unordered_map<string, vector<int>> m;
        for (int i = 0; i < words.size(); ++i) {
            m[words[i]].push_back(i);
        }
        int res = INT_MAX;
        if (word1 != word2) {
            for (int i : m[word1]) {
                for (int j : m[word2]) {
                    res = min(res, abs(i - j));
                }
            }
        } else {
            for (int i = 1; i < m[word1].size(); ++i) {
                res = min(res, m[word1][i] - m[word1][i - 1]);
            }
        }
        return res;
    }
};

日期

2019 年 9 月 24 日 —— 梦见回到了小学，小学已经芳草萋萋破败不堪

posted @ 2019-09-24 22:43  负雪明烛  阅读(35)  评论(0)  编辑  收藏  举报