243. Shortest Word Distance

题目：

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.

链接： http://leetcode.com/problemset/algorithms/

题解：

找两个单词在数组中的距离。这个题比较没有意义...就跟找两个字母的距离一样。乍看题目还以为是word ladder。设置两个变量，分别代表每个单词的lastIndex，之后就是判断和计算了。也可以用一个变量来记录。

Time Complexity - O(n)， Space Complexity - O(1)

public class Solution {
    public int shortestDistance(String[] words, String word1, String word2) {
        if(words == null || words.length == 0 || word1 == null || word2 == null)
            return Integer.MAX_VALUE;
        int minDistance = Integer.MAX_VALUE;
        int word1Index = -1, word2Index = -2;
        
        for(int i = 0; i < words.length; i++) {
            if(words[i].equals(word1))
                word1Index = i;
            if(words[i].equals(word2))
                word2Index = i;
            if(word1Index >= 0 && word2Index >= 0)
                minDistance = Math.min(minDistance, Math.abs(word1Index - word2Index));
        }
        
        return minDistance;
    }
}

 

二刷：

使用两个变量存下 word1和word2在数组中的位置，然后进行计算。

Java:

Time Complexity - O(n)， Space Complexity - O(1)

public class Solution {
    public int shortestDistance(String[] words, String word1, String word2) {
        if (words == null || words.length < 2 || word1 == null || word2 == null) {
            return Integer.MAX_VALUE;
        }
        int word1Pos = -1, word2Pos = -1, minDistance = Integer.MAX_VALUE;
        for (int i = 0; i < words.length; i++) {
            String curWord = words[i];
            if (curWord.equals(word1) || curWord.equals(word2)) {
                if (curWord.equals(word1)) {
                    word1Pos = i;
                }
                if (curWord.equals(word2)) {
                    word2Pos = i;
                }
                if (word1Pos >= 0 && word2Pos >= 0) {
                    minDistance = Math.min(minDistance, Math.abs(word1Pos - word2Pos));        
                }
            }
        }
        return minDistance;
    }
}

三刷：

Java:

public class Solution {
    public int shortestDistance(String[] words, String word1, String word2) {
        int minDist = Integer.MAX_VALUE;
        int word1Idx = -1, word2Idx = -1;
        for (int i = 0; i < words.length; i++) {
            if (words[i].equals(word1)) {
                word1Idx = i;
                if (word2Idx >= 0) minDist = Math.min(minDist, i - word2Idx);
            } else if (words[i].equals(word2)) {
                word2Idx = i;
                if (word1Idx >= 0) minDist = Math.min(minDist, i - word1Idx);
            }
        }
        return minDist;
    }
}

Reference: