class Solution {
public:
int shortestDistance(vector<string>& wordsDict, string word1, string word2) {
}
};
243. 最短单词距离
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 * 1041 <= wordsDict[i].length <= 10wordsDict[i] 由小写英文字母组成word1 和 word2 在 wordsDict 中word1 != word2原站题解
golang 解法, 执行用时: 0 ms, 内存消耗: 4 MB, 提交时间: 2023-10-15 18:58:37
func shortestDistance(wordsDict []string, word1, word2 string) int {
ans := len(wordsDict)
index1, index2 := -1, -1
for i, word := range wordsDict {
if word == word1 {
index1 = i
} else if word == word2 {
index2 = i
}
if index1 >= 0 && index2 >= 0 {
ans = min(ans, abs(index1-index2))
}
}
return ans
}
func abs(x int) int {
if x < 0 {
return -x
}
return x
}
func min(a, b int) int {
if a > b {
return b
}
return a
}
javascript 解法, 执行用时: 68 ms, 内存消耗: 42.1 MB, 提交时间: 2023-10-15 18:58:24
/**
* @param {string[]} wordsDict
* @param {string} word1
* @param {string} word2
* @return {number}
*/
var shortestDistance = function(wordsDict, word1, word2) {
const length = wordsDict.length;
let ans = length;
let index1 = -1, index2 = -1;
for (let i = 0; i < length; i++) {
const word = wordsDict[i];
if (word === word1) {
index1 = i;
} else if (word === word2) {
index2 = i;
}
if (index1 >= 0 && index2 >= 0) {
ans = Math.min(ans, Math.abs(index1 - index2));
}
}
return ans;
};
java 解法, 执行用时: 2 ms, 内存消耗: 43.5 MB, 提交时间: 2023-10-15 18:58:13
class Solution {
public int shortestDistance(String[] wordsDict, String word1, String word2) {
int length = wordsDict.length;
int ans = length;
int index1 = -1, index2 = -1;
for (int i = 0; i < length; i++) {
String word = wordsDict[i];
if (word.equals(word1)) {
index1 = i;
} else if (word.equals(word2)) {
index2 = i;
}
if (index1 >= 0 && index2 >= 0) {
ans = Math.min(ans, Math.abs(index1 - index2));
}
}
return ans;
}
}
cpp 解法, 执行用时: 4 ms, 内存消耗: 11.6 MB, 提交时间: 2023-10-15 18:58:00
class Solution {
public:
int shortestDistance(vector<string>& wordsDict, string word1, string word2) {
int length = wordsDict.size();
int ans = length;
int index1 = -1, index2 = -1;
for (int i = 0; i < length; i++) {
if (wordsDict[i] == word1) {
index1 = i;
} else if (wordsDict[i] == word2) {
index2 = i;
}
if (index1 >= 0 && index2 >= 0) {
ans = min(ans, abs(index1 - index2));
}
}
return ans;
}
};
python3 解法, 执行用时: 44 ms, 内存消耗: 18.8 MB, 提交时间: 2023-10-15 18:57:48
class Solution:
def shortestDistance(self, wordsDict: List[str], word1: str, word2: str) -> int:
ans = len(wordsDict)
index1, index2 = -1, -1
for i, word in enumerate(wordsDict):
if word == word1:
index1 = i
elif word == word2:
index2 = i
if index1 >= 0 and index2 >= 0:
ans = min(ans, abs(index1 - index2))
return ans