243. Shortest Word Distance

问题描述

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

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

Given word1 = “coding”, word2 = “practice”, return 3.
Given word1 = "makes", word2 = "coding", return 1.

Note:
You may assume that word1 does not equal to word2, and word1 and word2 are both in the list.

题目链接：

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思路分析

给一个string的数组，里面有word1和word2两个词，要求返回这两个词之间的最短距离。

使用两个指针，分别记录两个词的位置，差的绝对值的最小值就是最短距离。

指针的位置开始时初始化为-1，当两个指针都有有效值之后再开始判断。

代码

class Solution {
public:
    int shortestDistance(vector<string>& words, string word1, string word2) {
        int distance, temp, w1 = -1, w2 = -1;
        distance = words.size() - 1;
        for (int i = 0; i < words.size(); i++){
            if (words[i] ==  word1)
                w1 = i;
            if (words[i] ==  word2)
                w2 = i;
            if (w1 != -1 && w2 != -1){
                temp = abs(w1 - w2);
                if (distance > temp)
                    distance = temp;
            }
        }
        return distance;
    }
};

时间复杂度：$O(n)$
空间复杂度：$O(1)$

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反思

关于有多个word1和word2的情况，最短的距离肯定是存在在离的近的一对之间的，所以没有必要全部比较多。

感觉可以通过使用更高端的数据结构提升速度。