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Commit 3e3d06c

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feat: add solutions to lc problem: No.0245 (#4433)
1 parent ec22c87 commit 3e3d06c

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3 files changed

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-6
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‎solution/0200-0299/0245.Shortest Word Distance III/README.md‎

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### 方法一:分情况讨论
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先判断 `word1` 和 `word2` 是否相等:
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我们首先判断 $\textit{word1}$ 和 $\textit{word2}$ 是否相等:
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如果相等,遍历数组 `wordsDict`,找到两个 `word1` 的下标 $i$ 和 $j,ドル求 $i-j$ 的最小值。
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- 如果相等,遍历数组 $\textit{wordsDict},ドル找到两个 $\textit{word1}$ 的下标 $i$ 和 $j,ドル求 $i-j$ 的最小值。
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- 如果不相等,遍历数组 $\textit{wordsDict},ドル找到 $\textit{word1}$ 和 $\textit{word2}$ 的下标 $i$ 和 $j,ドル求 $i-j$ 的最小值。
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如果不相等,遍历数组 `wordsDict`,找到 `word1` 和 `word2` 的下标 $i$ 和 $j,ドル求 $i-j$ 的最小值。
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时间复杂度 $O(n),ドル空间复杂度 $O(1)$。其中 $n$ 为数组 `wordsDict` 的长度。
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时间复杂度 $O(n),ドル其中 $n$ 为数组 $\textit{wordsDict}$ 的长度。空间复杂度 $O(1)$。
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<!-- tabs:start -->
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}
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```
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#### TypeScript
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```ts
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function shortestWordDistance(wordsDict: string[], word1: string, word2: string): number {
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let ans = wordsDict.length;
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if (word1 === word2) {
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let j = -1;
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for (let i = 0; i < wordsDict.length; i++) {
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if (wordsDict[i] === word1) {
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if (j !== -1) {
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ans = Math.min(ans, i - j);
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}
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j = i;
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}
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}
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} else {
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let i = -1,
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j = -1;
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for (let k = 0; k < wordsDict.length; k++) {
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if (wordsDict[k] === word1) {
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i = k;
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}
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if (wordsDict[k] === word2) {
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j = k;
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}
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if (i !== -1 && j !== -1) {
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ans = Math.min(ans, Math.abs(i - j));
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}
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}
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}
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return ans;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->

‎solution/0200-0299/0245.Shortest Word Distance III/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: Case Analysis
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First, we check whether $\textit{word1}$ and $\textit{word2}$ are equal:
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- If they are equal, iterate through the array $\textit{wordsDict}$ to find two indices $i$ and $j$ of $\textit{word1},ドル and compute the minimum value of $i-j$.
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- If they are not equal, iterate through the array $\textit{wordsDict}$ to find the indices $i$ of $\textit{word1}$ and $j$ of $\textit{word2},ドル and compute the minimum value of $i-j$.
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The time complexity is $O(n),ドル where $n$ is the length of the array $\textit{wordsDict}$. The space complexity is $O(1)$.
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<!-- tabs:start -->
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}
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```
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#### TypeScript
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```ts
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function shortestWordDistance(wordsDict: string[], word1: string, word2: string): number {
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let ans = wordsDict.length;
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if (word1 === word2) {
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let j = -1;
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for (let i = 0; i < wordsDict.length; i++) {
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if (wordsDict[i] === word1) {
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if (j !== -1) {
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ans = Math.min(ans, i - j);
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}
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j = i;
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}
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}
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} else {
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let i = -1,
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j = -1;
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for (let k = 0; k < wordsDict.length; k++) {
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if (wordsDict[k] === word1) {
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i = k;
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}
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if (wordsDict[k] === word2) {
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j = k;
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}
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if (i !== -1 && j !== -1) {
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ans = Math.min(ans, Math.abs(i - j));
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}
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}
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}
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return ans;
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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function shortestWordDistance(wordsDict: string[], word1: string, word2: string): number {
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let ans = wordsDict.length;
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if (word1 === word2) {
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let j = -1;
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for (let i = 0; i < wordsDict.length; i++) {
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if (wordsDict[i] === word1) {
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if (j !== -1) {
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ans = Math.min(ans, i - j);
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}
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j = i;
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}
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}
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} else {
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let i = -1,
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j = -1;
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for (let k = 0; k < wordsDict.length; k++) {
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if (wordsDict[k] === word1) {
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i = k;
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}
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if (wordsDict[k] === word2) {
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j = k;
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}
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if (i !== -1 && j !== -1) {
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ans = Math.min(ans, Math.abs(i - j));
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}
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}
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}
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return ans;
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}

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