README_EN.md

243. Shortest Word Distance

中文文档

Description

Given an array of strings wordsDict and two different strings that already exist in the array word1 and word2, return the shortest distance between these two words in the list.

 

Example 1:

Input: wordsDict = ["practice", "makes", "perfect", "coding", "makes"], word1 = "coding", word2 = "practice"
Output: 3

Example 2:

Input: wordsDict = ["practice", "makes", "perfect", "coding", "makes"], word1 = "makes", word2 = "coding"
Output: 1

 

Constraints:

• 1 <= wordsDict.length <= 3 * 104
• 1 <= wordsDict[i].length <= 10
• wordsDict[i] consists of lowercase English letters.
• word1 and word2 are in wordsDict.
• word1 != word2

Solutions

Python3

class Solution:
    def shortestDistance(self, wordsDict: List[str], word1: str, word2: str) -> int:
        i1 = i2 = -1
        shortest_distance = len(wordsDict)
        for i in range(len(wordsDict)):
            if wordsDict[i] == word1:
                i1 = i
            elif wordsDict[i] == word2:
                i2 = i
            if i1 != -1 and i2 != -1:
                shortest_distance = min(shortest_distance, abs(i1 - i2))
        return shortest_distance

Java

class Solution {
    public int shortestDistance(String[] wordsDict, String word1, String word2) {
        int i1 = -1, i2 = -1;
        int shortestDistance = wordsDict.length;
        for (int i = 0; i < wordsDict.length; ++i) {
            if (word1.equals(wordsDict[i])) {
                i1 = i;
            } else if (word2.equals(wordsDict[i])) {
                i2 = i;
            }
            if (i1 != -1 && i2 != -1) {
                shortestDistance = Math.min(shortestDistance, Math.abs(i1 - i2));
            }
        }
        return shortestDistance;
    }
}

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