[leetcode 245] Shortest Word Distance III

Question:

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.

分析：

求两个单词在列表中的最短距离，这两个单词有可能相同，循环一遍依次判断即可。用p1表示word1的位置，用p2表示word2的位置。每次返回最小的距离即可。

直接看代码。

代码如下：

<pre name="code" class="cpp"><span style="font-size:14px;">class Solution {
public:
int shortestWordDistance(vector<string>& words, string word1, string word2) {
int n = words.size();
int p1 = -1, p2 = -1, dist = INT_MAX;
for(int i=0; i<n; ++i){
if(word1==word2){
if(words[i]==word1){
if(p1 == -1) p1 = i;
else
if(p2 == -1)
p1 = i;
else{
p1 = p2;
p2 = i;
}
}
}else{
if(words[i]==word1) p1 = i;
if(words[i]==word2) p2 = i;
}
if(p1>=0 && p2>=0)
dist = min(dist, abs(p1-p2));
}
return dist;
}
};</span>