LeetCode 245. Shortest Word Distance III

This is a follow up of Shortest Word Distance. The only difference is now word1 could be the same as word2.

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.

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

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

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

思路：
1. 和243比，要处理word1和word2相同的case。当相同的时候，怎么处理？
2. 用p1代表word1所在的位置，p2代表word2所在位置。实时更新p1和p2，如果word1==word2,则把新的位置赋给p2, 把p2赋给p1。

int shortestWordDistance(vector<string>& words, string word1, string word2) {    int p1=words.size(),p2=-words.size();//赋这样的初值是保证开始p1-p2长度超过words.size();    int res=INT_MAX;    for(int i=0;i<words.size();i++){        if(words[i]==word1)            p1=(word1==word2)?p2:i;        if(words[i]==word2)            p2=i;        res=min(res,abs(p1-p2));    }}