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.
此题开始的时候没有想出直接用数组该怎么做，后来用了hashmap做的，通过了：
public class Solution {
public int shortestDistance(String[] words, String word1, String word2) {
int min = Integer.MAX_VALUE;
Map<String,Integer> map =new HashMap<>();
int w1 = Integer.MIN_VALUE;
int w2 = Integer.MIN_VALUE;
for(int i=0;i<words.length;i++){
map.put(words[i],i);
if(map.containsKey(word1)){
w1 = map.get(word1);
}
if(map.containsKey(word2)){
w2 = map.get(word2);
}
if(w1!=Integer.MIN_VALUE&&w2!=Integer.MIN_VALUE){
min = Math.min(min,Math.abs(w1-w2));
}
}
return min==Integer.MAX_VALUE?-1:min;
}
}
看了答案，直接用数组就可以直接解出来，代码如下：
public class Solution {
public int shortestDistance(String[] words, String word1, String word2) {
int min = Integer.MAX_VALUE;
int p1 = -1;
int p2 = -1;
for(int i=0;i<words.length;i++){
if(words[i].equals(word1)) p1 = i;
if(words[i].equals(word2)) p2 = i;
if(p1!=-1&&p2!=-1){
min = Math.min(min,Math.abs(p1-p2));
}
}
return min;
}
}
这时双指针的题目，但是不清楚是不是隶属于greedy题。