977. Squares of a Sorted Array C++

原创于 2022-03-02 11:17:18 发布 · 201 阅读

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本文介绍如何在O(n)时间内，对于已排序的整数数组，计算每个元素的平方并保持非递减顺序。通过不同的算法思路，挑战传统排序后再平方的方法，提供一种更高效的解决方案。

Given an integer array nums sorted in non-decreasing order, return an array of the squares of each number sorted in non-decreasing order.

Example 1:

Input: nums = [-4,-1,0,3,10]
Output: [0,1,9,16,100]
Explanation: After squaring, the array becomes [16,1,0,9,100].
After sorting, it becomes [0,1,9,16,100].

Example 2:

Input: nums = [-7,-3,2,3,11]
Output: [4,9,9,49,121]

Constraints:

1 <= nums.length <= 104
-104 <= nums[i] <= 104
nums is sorted in non-decreasing order.

Follow up: Squaring each element and sorting the new array is very trivial, could you find an O(n) solution using a different approach?

class Solution {
public:
    vector<int> sortedSquares(vector<int>& nums) {
        vector<int> res;
        for (int num:nums){
            res.push_back(num*num);
        }
        sort(res.begin(),res.end());
        return res;
    }
};