[LeetCode] 243. Shortest Word Distance 最短单词距离
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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.
给一个单词数组和两个单词,返回这两个单词在数组里的最短距离。假定两个单词不同,而且都在数组中。
Java:
public int shortestDistance(String[] words, String word1, String word2) { int m=-1; int n=-1; int min = Integer.MAX_VALUE; for(int i=0; i<words.length; i++){ String s = words[i]; if(word1.equals(s)){ m = i; if(n!=-1) min = Math.min(min, m-n); }else if(word2.equals(s)){ n = i; if(m!=-1) min = Math.min(min, n-m); } } return min; }
Python:
# Time: O(n) # Space: O(1) class Solution: # @param {string[]} words # @param {string} word1 # @param {string} word2 # @return {integer} def shortestDistance(self, words, word1, word2): dist = float("inf") i, index1, index2 = 0, None, None while i < len(words): if words[i] == word1: index1 = i elif words[i] == word2: index2 = i if index1 is not None and index2 is not None: dist = min(dist, abs(index1 - index2)) i += 1 return dist
C++:
class Solution { public: int shortestDistance(vector<string>& words, string word1, string word2) { int p1 = -1, p2 = -1, res = INT_MAX; for (int i = 0; i < words.size(); ++i) { if (words[i] == word1) p1 = i; else if (words[i] == word2) p2 = i; if (p1 != -1 && p2 != -1) res = min(res, abs(p1 - p2)); } return res; } };
C++:
class Solution { public: int shortestDistance(vector<string>& words, string word1, string word2) { int idx = -1, res = INT_MAX; for (int i = 0; i < words.size(); ++i) { if (words[i] == word1 || words[i] == word2) { if (idx != -1 && words[idx] != words[i]) { res = min(res, i - idx); } idx = i; } } return res; } };
类似题目:
[LeetCode] 244. Shortest Word Distance II 最短单词距离 II
[LeetCode] 245. Shortest Word Distance III 最短单词距离 III
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