问题描述
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.
解决思路
- 循环查找比较的word1和word2，分别记录下index1和index2
- 比较index1和index2之差的绝对值，循环计算最小distance替换min
代码
package leetcode;
public class ShortestWordDistance {
public int shortestDistance(String[] words, String word1, String word2) {
int index1 = 0;
int index2 = 0;
int dis = 0;
int min = Integer.MAX_VALUE;
for (int i = 0; i < words.length; i++) {
if (words[i].equals(word1)){
index1 = i;
for (int j = 0; j < words.length; j++) {
if (words[j].equals(word2)){
index2 = j;
//calculate the distance
dis = Math.abs(i-j);
//find the min distance
if (dis < min){
min = dis;
}
}
}
}
}
return min;
}
}
代码改进
package leetcode;
public class ShortestWordDistance {
public int shortestDistance(String[] words, String word1, String word2) {
int index1 = -1;
int index2 = -1;
int min = Integer.MAX_VALUE;
for (int i = 0; i < words.length; i++) {
if (words[i].equals(word1)) index1 = i;
if (words[i].equals(word2)) index2 = i;
if(index1 != -1 && index2 != -1){
min = Math.min(min, Math.abs(index1 - index2));
}
}
return min;
}
}
小感想
注意：比较两个非基本数据类型的值的时候，要用equals函数比较，不能用==，因为==比较的是内存地址，而equals才是比较的内容。在这里用==，虽然结果没错（因为在底层存储数组的时候，比如说words数组之前有一个”practic”，然后再次存入”practice”,Java会从缓冲区查找是否有该值，有的话会将该地址直接赋给第二个”practice”，所以这样做，如果两个值一样，地址也是一样的），但是原理上来讲，String[]不是基本数据类型，这样是比较在地址值，是不对的！！！
