Question

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
}