[LeetCode] 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?

有序数组的平方。

给你一个按 非递减顺序 排序的整数数组 nums，返回 每个数字的平方 组成的新数组，要求也按 非递减顺序 排序。

思路

思路是 two pointer 夹逼。创建一个跟 input 同样长度的数组。因为 input 里面存在负数，而且很有可能存在一个绝对值非常大的负数，导致其平方值最大，所以需要 two pointer 左右指针比较哪个数字的绝对值大。细节实现参见代码。

复杂度

时间O(n)
空间O(n)

代码

Java实现

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

JavaScript实现

/**
 * @param {number[]} A
 * @return {number[]}
 */
var sortedSquares = function (A) {
    let n = A.length;
    let res = [n];
    let i = 0;
    let j = n - 1;
    for (let p = n - 1; p >= 0; p--) {
        if (Math.abs(A[i]) > Math.abs(A[j])) {
            res[p] = A[i] * A[i];
            i++;
        } else {
            res[p] = A[j] * A[j];
            j--;
        }
    }
    return res;
};