245. Shortest Word Distance III 🔒

中文文档

Description

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

Note that word1 and word2 may be the same. It is guaranteed that they represent two individual words in the list.

 

Example 1:

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

Example 2:

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

 

Constraints:

• 1 <= wordsDict.length <= 105
• 1 <= wordsDict[i].length <= 10
• wordsDict[i] consists of lowercase English letters.
• word1 and word2 are in wordsDict.

Solutions

Solution 1

Python3

class Solution:
    def shortestWordDistance(self, wordsDict: List[str], word1: str, word2: str) -> int:
        ans = len(wordsDict)
        if word1 == word2:
            j = -1
            for i, w in enumerate(wordsDict):
                if w == word1:
                    if j != -1:
                        ans = min(ans, i - j)
                    j = i
        else:
            i = j = -1
            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

Java

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

C++

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

Go

func shortestWordDistance(wordsDict []string, word1 string, word2 string) int {
	ans := len(wordsDict)
	if word1 == word2 {
		j := -1
		for i, w := range wordsDict {
			if w == word1 {
				if j != -1 {
					ans = min(ans, i-j)
				}
				j = i
			}
		}
	} else {
		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
}