103 lines
1.9 KiB
Markdown
103 lines
1.9 KiB
Markdown
# Leetcode Squares-of-a-Sorted-Array
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#### 2022-07-10 08:50
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> ##### Algorithms:
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>
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> #algorithm #two_pointers
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>
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> ##### Data structures:
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>
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> #DS #array
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>
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> ##### Difficulty:
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>
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> #coding_problem #difficulty_easy
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>
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> ##### Additional tags:
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>
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> #leetcode
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>
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> ##### Revisions:
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>
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> N/A
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##### Related topics:
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##### Links:
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- [Link to problem](https://leetcode.com/problems/squares-of-a-sorted-array/)
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---
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### Problem
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Given an integer array `nums` sorted in **non-decreasing** order, return _an array of **the squares of each number** sorted in non-decreasing order_.
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Follow up: Squaring each element and sorting the new array is very trivial, could you find an O(n) solution using a different approach?
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#### Examples
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**Example 1:**
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**Input:** nums = [-4,-1,0,3,10]
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**Output:** [0,1,9,16,100]
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**Explanation:** After squaring, the array becomes [16,1,0,9,100].
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After sorting, it becomes [0,1,9,16,100].
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**Example 2:**
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**Input:** nums = [-7,-3,2,3,11]
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**Output:** [4,9,9,49,121]
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#### Constraints
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- `1 <= nums.length <= 104`
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- `-104 <= nums[i] <= 104`
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- `nums` is sorted in **non-decreasing** order.
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### Thoughts
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> [!summary]
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> This is a #two_pointers
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Using two pointers.
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One from left, one from right end.
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**fill the array in backwards order.**
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### Solution
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cpp
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```cpp
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class Solution {
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public:
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vector<int> sortedSquares(vector<int> &nums) {
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// Double pointers, since array is sorted and access in order.
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// O(2n)
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int mid;
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int size = nums.size();
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int left = 0, right = size - 1;
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int count = size - 1;
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vector<int> answer(size);
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for (int i = 0; i < size; i++) {
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nums[i] = nums[i] * nums[i];
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}
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while (left <= right) {
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if (nums[left] >= nums[right]) {
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answer[count--] = nums[left];
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left++;
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} else if (nums[right] > nums[left]) {
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answer[count--] = nums[right];
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right--;
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}
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}
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return answer;
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}
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};
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```
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