題目地址:
https://www.lintcode.com/problem/shortest-word-distance/description
給定一個字符串數組,再給定兩個單詞,求這兩個單詞在字符串中最小的下標距離(數組中可能含有某個單詞多次)。
用兩個指針記錄兩個單詞出現的位置,然後不斷更新距離即可。代碼如下:
public class Solution {
/**
* @param words: a list of words
* @param word1: a string
* @param word2: a string
* @return: the shortest distance between word1 and word2 in the list
*/
public int shortestDistance(String[] words, String word1, String word2) {
// Write your code here
int idx1 = -1, idx2 = -1;
int res = words.length;
for (int i = 0; i < words.length; i++) {
if (words[i].equals(word1)) {
idx1 = i;
}
if (words[i].equals(word2)) {
idx2 = i;
}
// 如果兩個字符串都找到了,則算一下距離,並更新
if (idx1 != -1 && idx2 != -1) {
res = Math.min(res, Math.abs(idx1 - idx2));
}
}
return res;
}
}
時間複雜度,空間。