997. Square of ordered array (array)

subject ：

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 .

Example 1： Input ：nums = [-4,-1,0,3,10] Output ：[0,1,9,16,100] explain ： After square , The array becomes [16,1,0,9,100], After ordering , The array becomes [0,1,9,16,100]

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

Method 1 ： Violence solution

import java.util.Arrays;

public class Youxushuzupingfang {
    public int[] sortedSquares(int [] nums){
        for (int i = 0; i <nums.length ; i++) {
            nums[i]*=nums[i];
        }
        // Quick line up 
        Arrays.sort(nums);
        return nums;
    }

    public static void main(String[] args) {
        int [] nums = {-7,-3,2,3,11};
        Youxushuzupingfang youxushuzupingfang = new Youxushuzupingfang();
        int [] res =youxushuzupingfang.sortedSquares(nums);
        for (int x:res) {
            System.out.print(x+" ");
        }
    }
}

Time complexity ：O（n+nlogn）： among n Is the first for Generated by cyclic square ,nlogn It is generated by quick sort

Spatial complexity ：O（logn）

Output results ：

4 9 9 49 121 

Method 2 ： Double finger needling

The maximum square can only be generated at both ends of the array , Can't be in the middle of an array , Therefore, two finger acupuncture can be used , Point to the left and right ends of the array , Compare the size of the square of the number pointed to by the pointer .

public int[] sortedSquares(int [] nums){
        int n = nums.length;
        // Create a new array and save the squared value 
        int [] temp = new int [n];
        // Set the left and right pointers 
        for (int i=0,j=n-1,k=n-1; k>=0 ;k--) {
            if(nums[i]*nums[i]>nums[j]*nums[j]){
                temp[k] = nums[i]*nums[i];
                i++;
            }else if(nums[i]*nums[i]<nums[j]*nums[j]){
                temp[k] = nums[j]*nums[j];
                j--;
            }else {//i,j The value after the square of the element pointed to is equal 
                temp[k] = nums[i]*nums[i];
            }
        }
        return temp;
    }

    public static void main(String[] args) {
        int [] nums = {-7,-3,2,3,11};
        Youxushuzupingfang youxushuzupingfang = new Youxushuzupingfang();
        int [] res =youxushuzupingfang.sortedSquares(nums);
        for (int x:res) {
            System.out.print(x+" ");
        }
    }

Time complexity ：O（n）

Spatial complexity ：O（n）

Other versions of the double pointer solution ： Compare the code above , There are some things worth improving , Use while The cycle does not need to judge the number of cycles , Left and right pointers use left,right More intuitive expression

class Solution {
    public int[] sortedSquares(int[] nums) {
        int right = nums.length - 1;
        int left = 0;
        int[] result = new int[nums.length];
        int index = result.length - 1;
        while (left <= right) {
            if (nums[left] * nums[left] > nums[right] * nums[right]) {
                result[index--] = nums[left] * nums[left];
                ++left;
            } else {
                result[index--] = nums[right] * nums[right];
                --right;
            }
        }
        return result;
    }
}