[LeetCode] 243. Shortest Word Distance 最短单词距离

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

原文地址：https://www.cnblogs.com/lightwindy/p/9736289.html