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243. 最短单词距离

English Version

题目描述

给定一个字符串数组 wordDict 和两个已经存在于该数组中的不同的字符串 word1 和 word2 。返回列表中这两个单词之间的最短距离。

 

示例 1:

输入: wordsDict = ["practice", "makes", "perfect", "coding", "makes"], word1 = "coding", word2 = "practice"
输出: 3

示例 2:

输入: wordsDict = ["practice", "makes", "perfect", "coding", "makes"], word1 = "makes", word2 = "coding"
输出: 1

 

提示:

• 1 <= wordsDict.length <= 3 * 104
• 1 <= wordsDict[i].length <= 10
• wordsDict[i] 由小写英文字母组成
• word1 和 word2 在 wordsDict 中
• word1 != word2

解法

方法一：双指针

遍历数组 wordsDict，找到 word1 和 word2 的下标 $i$ 和 $j$，求 $i-j$ 的最小值。

时间复杂度 $O(n)$，空间复杂度 $O(1)$。其中 $n$ 为数组 wordsDict 的长度。

class Solution:
    def shortestDistance(self, wordsDict: List[str], word1: str, word2: str) -> int:
        i = j = -1
        ans = inf
        for k, w in enumerate(wordsDict):
            if w == word1:
                i = k
            if w == word2:
                j = k
            if i != -1 and j != -1:
                ans = min(ans, abs(i - j))
        return ans

class Solution {
    public int shortestDistance(String[] wordsDict, String word1, String word2) {
        int ans = 0x3f3f3f3f;
        for (int k = 0, i = -1, j = -1; k < wordsDict.length; ++k) {
            if (wordsDict[k].equals(word1)) {
                i = k;
            }
            if (wordsDict[k].equals(word2)) {
                j = k;
            }
            if (i != -1 && j != -1) {
                ans = Math.min(ans, Math.abs(i - j));
            }
        }
        return ans;
    }
}

class Solution {
public:
    int shortestDistance(vector<string>& wordsDict, string word1, string word2) {
        int ans = INT_MAX;
        for (int k = 0, i = -1, j = -1; k < wordsDict.size(); ++k) {
            if (wordsDict[k] == word1) {
                i = k;
            }
            if (wordsDict[k] == word2) {
                j = k;
            }
            if (i != -1 && j != -1) {
                ans = min(ans, abs(i - j));
            }
        }
        return ans;
    }
};

func shortestDistance(wordsDict []string, word1 string, word2 string) int {
	ans := 0x3f3f3f3f
	i, j := -1, -1
	for k, w := range wordsDict {
		if w == word1 {
			i = k
		}
		if w == word2 {
			j = k
		}
		if i != -1 && j != -1 {
			ans = min(ans, abs(i-j))
		}
	}
	return ans
}

func abs(x int) int {
	if x < 0 {
		return -x
	}
	return x
}