【LeetCode】245. Shortest Word Distance III 解题报告 (C++)
- 作者: 负雪明烛
- id: fuxuemingzhu
- 个人博客:http://fuxuemingzhu.cn/
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:
- 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。
- 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
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