Shortest Word Distance
題意：
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.
解題思路：
updated 2016.1.7
較naive，但程式較複雜點，為什麼找到了還要繼續判斷而不直接返回，是因為陣列中面能含有兩個以上相同的字串。
public class Solution {
public int shortestDistance(String[] words, String word1, String word2) {
if (words == null || words.length < 2) {
return 0;
}
int idxOne = -1;
int idxTwo = -1;
int minLength = Integer.MAX_VALUE;
for (int i = 0; i < words.length; i++) {
if (words[i].equals(word1)) {
idxOne = i;
} else if (words[i].equals(word2)) {
idxTwo = i;
}
// 為什麼找到了還要繼續判斷而不直接返回，是因為陣列中面能含有兩個以上相同的字串
if (idxOne != -1 && idxTwo != -1) {
minLength = Math.min(minLength, Math.abs(idxOne - idxTwo));
}
}
return minLength;
}
}
只使用一個index來標記在當下字串之前遇到的另一個字的index為多少，記得再判斷index指的那個字與當下的字不同，再更新mindistance。
public class Solution {
public int shortestDistance(String[] words, String word1, String word2) {
int index = -1;
int minDistance = Integer.MAX_VALUE;
for (int i = 0; i < words.length; i++) {
if (words[i].equals(word1) || words[i].equals(word2)) {
if (index != -1 && !words[index].equals(words[i])) {
minDistance = Math.min(minDistance, i - index);
}
index = i;
}
}
return minDistance;
}
}