Force buckle 977 Square of ordered array

Here's a button   Non decreasing order   Sorted array of integers  nums, return   The square of each number   A new array of , According to the requirements   Non decreasing order   Sort .

Method ： Double pointer , Point to the beginning and end of the array respectively  

class Solution {
public:
    /* The following method is simple , But not recommended , The time complexity is  O(n + nlogn)*/
    // vector<int> sortedSquares(vector<int>& nums) {
    //     for(int i = 0; i < nums.size(); ++i) {
    //         nums[i] *= nums[i];
    //     }
    //     sort(nums.begin(),nums.end());  Quick sort 
    //     return nums;
    // }

    /* Double finger needling , Because the array is arranged in ascending order , The largest number after square must be either on the left or on the right , It can't be in the middle 
     So let's compare the squares from both sides */
    vector<int> sortedSquares(vector<int>& nums) {
        vector<int> result(nums.size(), 0);
        int k = nums.size() - 1;
        //i Point to the beginning of the array ,j Point to the end of the array 
        for(int i = 0, j = nums.size() - 1; i <= j;) { //i <= j, To consider i and j There is an element in the middle 
            if(nums[i] * nums[i] < nums[j] * nums[j]) {
                result[k--] = nums[j] * nums[j];
                --j;
            }
            else{
                result[k--] = nums[i] * nums[i];
                ++i;
            }
        }
        return result;
    }
    // Time complexity ：O(n)
};