977. Squares of a Sorted Array

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?

Solution:

class Solution {
public:
    vector<int> sortedSquares(vector<int>& nums) {
        vector<int> result(nums.size());
        int le = 0;
        int ri = nums.size()-1;
        int length = nums.size()-1;
        while(le <= ri)
        {
            if(abs(nums[le]) > nums[ri])
            {
                result[length] = pow(abs(nums[le]),2);
                le++;
            }else{
                result[length] = pow(abs(nums[ri]),2);
                ri--;
            }
            length--;
        }
        return result;
    }
};

标签：977,le,nums,ri,length,result,Sorted,Array,array
来源： https://www.cnblogs.com/Pomelos/p/15773346.html