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Commit a1122bc

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feat: update solutions to lc problem: No.0219 (doocs#4001)
No.0219.Contains Duplicate II
1 parent f42c382 commit a1122bc

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3 files changed

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-24
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3 files changed

+21
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‎solution/0200-0299/0219.Contains Duplicate II/README.md‎

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@@ -60,13 +60,13 @@ tags:
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### 方法一:哈希表
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我们用哈希表 $d$ 存放最近遍历到的数以及对应的下标。
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我们用一个哈希表 $\textit{d}$ 存放最近遍历到的数以及对应的下标。
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遍历数组 `nums`,对于当前遍历到的元素 $nums[i],ドル如果在哈希表中存在,并且下标与当前元素的下标之差不超过 $k,ドル则返回 `true`,否则将当前元素加入哈希表中。
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遍历数组 $\textit{nums}$,对于当前遍历到的元素 $\textit{nums}[i],ドル如果在哈希表中存在,并且下标与当前元素的下标之差不超过 $k,ドル则返回 $\text{true}$,否则将当前元素加入哈希表中。
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遍历结束后,返回 `false`。
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遍历结束后,返回 $\text{false}$。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 为数组 `nums` 的长度。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。其中 $n$ 为数组 $\textit{nums}$ 的长度。
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<!-- tabs:start -->
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@@ -195,13 +195,12 @@ class Solution {
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* @return Boolean
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*/
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function containsNearbyDuplicate($nums, $k) {
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$hashtable = [];
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for ($i = 0; $i < count($nums); $i++) {
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$tmp = $nums[$i];
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if (array_key_exists($tmp, $hashtable) && $k >= $i - $hashtable[$tmp]) {
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$d = [];
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for ($i = 0; $i < count($nums); ++$i) {
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if (array_key_exists($nums[$i], $d) && $i - $d[$nums[$i]] <= $k) {
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return true;
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}
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$hashtable[$tmp] = $i;
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$d[$nums[$i]] = $i;
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}
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return false;
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}

‎solution/0200-0299/0219.Contains Duplicate II/README_EN.md‎

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@@ -59,13 +59,13 @@ tags:
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### Solution 1: Hash Table
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We use a hash table $d$ to store the nearest index of the number it has visited.
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We use a hash table $\textit{d}$ to store the recently traversed numbers and their corresponding indices.
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We traverse the array `nums`. For the current element $nums[i],ドル if it exists in the hash table, and the difference between its index and the current index is no larger than $k,ドル then return `true`. Otherwise, we add the current element into the hash table.
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Traverse the array $\textit{nums}$. For the current element $\textit{nums}[i],ドル if it exists in the hash table and the difference between the indices is no more than $k,ドル return $\text{true}$. Otherwise, add the current element to the hash table.
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After the traversal, return `false`.
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After traversing, return $\text{false}$.
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The time complexity is $O(n)$ and the space complexity is $O(n)$. Here $n$ is the length of array `nums`.
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The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $\textit{nums}$.
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<!-- tabs:start -->
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@@ -194,13 +194,12 @@ class Solution {
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* @return Boolean
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*/
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function containsNearbyDuplicate($nums, $k) {
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$hashtable = [];
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for ($i = 0; $i < count($nums); $i++) {
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$tmp = $nums[$i];
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if (array_key_exists($tmp, $hashtable) && $k >= $i - $hashtable[$tmp]) {
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$d = [];
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for ($i = 0; $i < count($nums); ++$i) {
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if (array_key_exists($nums[$i], $d) && $i - $d[$nums[$i]] <= $k) {
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return true;
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}
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$hashtable[$tmp] = $i;
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$d[$nums[$i]] = $i;
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}
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return false;
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}

‎solution/0200-0299/0219.Contains Duplicate II/Solution.php‎

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Original file line numberDiff line numberDiff line change
@@ -5,14 +5,13 @@ class Solution {
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* @return Boolean
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*/
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function containsNearbyDuplicate($nums, $k) {
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$hashtable = [];
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for ($i = 0; $i < count($nums); $i++) {
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$tmp = $nums[$i];
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if (array_key_exists($tmp, $hashtable) && $k >= $i - $hashtable[$tmp]) {
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$d = [];
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for ($i = 0; $i < count($nums); ++$i) {
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if (array_key_exists($nums[$i], $d) && $i - $d[$nums[$i]] <= $k) {
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return true;
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}
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$hashtable[$tmp] = $i;
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$d[$nums[$i]] = $i;
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}
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return false;
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}
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}
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}

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