LeetCode 245. Shortest Word Distance III（最短单词距离）

原题网址：https://leetcode.com/problems/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：对于word1==word2的情形单独处理，只需要用一个变量保存最近的位置。时间复杂度O(n)，空间复杂度O(1)

public class Solution {
public int shortestWordDistance(String[] words, String word1, String word2) {
if (word1.equals(word2)) {
int distance = Integer.MAX_VALUE;
int current = -1;
for(int i=0; i<words.length; i++) {
if (word1.equals(words[i])) {
if (current == -1) current = i;
else {
if (i-current < distance) distance = i-current;;
current = i;
}
}
}
return distance;
} else {
int distance = Integer.MAX_VALUE;
int pos1 = -1, pos2 = -1;
for(int i=0; i<words.length; i++) {
if (word1.equals(words[i])) {
pos1 = i;
if (pos2 != -1 && pos1-pos2<distance) distance = pos1-pos2;
} else if (word2.equals(words[i])) {
pos2 = i;
if (pos1 != -1 && pos2-pos1<distance) distance = pos2-pos1;
}
}
return distance;
}
}
}