题目描述
给定一个字符串数组 wordsDict 和两个字符串 word1 和 word2 ，返回这两个单词在列表中出现的最短距离。
注意：word1 和 word2 是有可能相同的，并且它们将分别表示为列表中 两个独立的单词 。
示例 1：
输入：wordsDict = ["practice", "makes", "perfect", "coding", "makes"], word1 = "makes", word2 = "coding"
输出：1
示例 2：
输入：wordsDict = ["practice", "makes", "perfect", "coding", "makes"], word1 = "makes", word2 = "makes"
输出：3
提示：
1 <= wordsDict.length <= 1051 <= wordsDict[i].length <= 10wordsDict[i] 由小写英文字母组成word1 和 word2 都在 wordsDict 中
解法
方法一：分情况讨论
我们首先判断 \(\textit{word1}\) 和 \(\textit{word2}\) 是否相等：
- 如果相等，遍历数组 \(\textit{wordsDict}\)，找到两个 \(\textit{word1}\) 的下标 \(i\) 和 \(j\)，求 \(i-j\) 的最小值。
- 如果不相等，遍历数组 \(\textit{wordsDict}\)，找到 \(\textit{word1}\) 和 \(\textit{word2}\) 的下标 \(i\) 和 \(j\)，求 \(i-j\) 的最小值。
时间复杂度 \(O(n)\)，其中 \(n\) 为数组 \(\textit{wordsDict}\) 的长度。空间复杂度 \(O(1)\)。
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20 | 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
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28 | 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;
}
}
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30 | 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;
}
};
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35 | 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
}
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29 | function shortestWordDistance(wordsDict: string[], word1: string, word2: string): number {
let ans = wordsDict.length;
if (word1 === word2) {
let j = -1;
for (let i = 0; i < wordsDict.length; i++) {
if (wordsDict[i] === word1) {
if (j !== -1) {
ans = Math.min(ans, i - j);
}
j = i;
}
}
} else {
let i = -1,
j = -1;
for (let k = 0; k < wordsDict.length; k++) {
if (wordsDict[k] === word1) {
i = k;
}
if (wordsDict[k] === word2) {
j = k;
}
if (i !== -1 && j !== -1) {
ans = Math.min(ans, Math.abs(i - j));
}
}
}
return ans;
}
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