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.
一开始想复杂了 总想直接找到最相近的两个索引
public int shortestDistance(String[] words, String word1, String word2) { List<Integer> list = new ArrayList<>(); TreeSet<Integer> set = new TreeSet<>(); for (int i=0; i<words.length; i++) { if (words[i].equals(word1)) { list.add(i); } else if (words[i].equals(word2)) { set.add(i); } } int min = Integer.MAX_VALUE; for (Integer n : list) { Integer small = set.floor(n); if (small != null) min = Math.min(min, n-small); Integer big = set.ceiling(n); if (big != null) min = Math.min(min, big-n); } return min; }实际上不需要这样 只要找出所有word的索引 遍历求差值就可以了
public int shortestDistance(String[] words, String word1, String word2) { int i1 = -1, i2 = -1; int minDistance = words.length; int currentDistance; for (int i = 0; i < words.length; i++) { if (words[i].equals(word1)) { i1 = i; } else if (words[i].equals(word2)) { i2 = i; } if (i1 != -1 && i2 != -1) { minDistance = Math.min(minDistance, Math.abs(i1 - i2)); } } return minDistance; }