87 lines
1.8 KiB
Markdown
87 lines
1.8 KiB
Markdown
# Leetcode Move-Zeroes
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#### 2022-07-11 09:55
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> ##### Algorithms:
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> #algorithm #two_pointers #array_in_place_operation
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> ##### Data structures:
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> #DS #array
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> ##### Difficulty:
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> #coding_problem #difficulty-easy
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> ##### Additional tags:
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> #leetcode
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> ##### Revisions:
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> N/A
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##### Related topics:
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```expander
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tag:#array_in_place_operation OR tag:#two_pointers
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```
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- [[Leetcode Intersection-of-Two-Arrays-II]]
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- [[Leetcode Merge-Sorted-Array]]
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- [[Leetcode Merge-Two-Sorted-Lists]]
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- [[Leetcode Rotate-Array]]
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- [[Leetcode Squares-of-a-Sorted-Array]]
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- [[Two pointers approach]]
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##### Links:
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- [Link to problem](https://leetcode.com/problems/move-zeroes/)
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___
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### Problem
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Given an integer array `nums`, move all `0`'s to the end of it while maintaining the relative order of the non-zero elements.
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**Note** that you must do this in-place without making a copy of the array.
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**Follow up:** Could you minimize the total number of operations done?
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#### Examples
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**Example 1:**
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**Input:** nums = [0,1,0,3,12]
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**Output:** [1,3,12,0,0]
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**Example 2:**
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**Input:** nums = [0]
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**Output:** [0]
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#### Constraints
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**Constraints:**
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- `1 <= nums.length <= 104`
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- `-231 <= nums[i] <= 231 - 1`
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### Thoughts
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> [!summary]
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> This is a #array_in_place_operation problem.
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Because the data can be overwritten, and zeros can be predicted in the result array, I use in-place array operations.
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### Solution
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```cpp
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class Solution {
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public:
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void moveZeroes(vector<int> &nums) {
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// Tow pointers, since data can be overwritten and one pointer is faster
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// than another.
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int num = 0;
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for (int i = 0; i < nums.size(); i++) {
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if (nums[i] != 0) {
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nums[num++] = nums[i];
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}
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}
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for (int j = num; j < nums.size(); j++) {
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nums[j] = 0;
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}
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}
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};
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``` |