LeetCode #977 Squares of a Sorted Array 有序数组的平方

977 Squares of a Sorted Array 有序数组的平方

Description:
Given an array of integers A sorted in non-decreasing order, return an array of the squares of each number, also in sorted non-decreasing order.

Example:

Example 1:

Input: [-4,-1,0,3,10]
Output: [0,1,9,16,100]

Example 2:

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

Note:

1 <= A.length <= 10000
-10000 <= A[i] <= 10000
A is sorted in non-decreasing order.

题目描述:
给定一个按非递减顺序排序的整数数组 A，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。

示例 :

示例 1：

输入：[-4,-1,0,3,10]
输出：[0,1,9,16,100]

示例 2：

输入：[-7,-3,2,3,11]
输出：[4,9,9,49,121]

提示：

1 <= A.length <= 10000
-10000 <= A[i] <= 10000
A 已按非递减顺序排序。

思路:

1. 直接原地平方之后再排序
   时间复杂度O(nlgn), 空间复杂度O(1)
2. 双指针法, 由于原数组已经有序, 依次从后往前添加平方值较大的元素即可
   时间复杂度O(n), 空间复杂度O(1)

代码:
C++:

class Solution 
{
public:
    vector sortedSquares(vector& A) 
    {
        int i = 0, j = A.size() - 1, k = A.size() - 1;
        vector result(A.size());
        while (i <= j)
        {
            if (A[i] * A[i] < A[j] * A[j]) result[k--] = A[j] * A[j--];
            else result[k--] = A[i] * A[i++];
        }
        return result;
    }
};

Java:

class Solution {
    public int[] sortedSquares(int[] A) {
        int i = 0, j = A.length - 1, k = A.length - 1, result[] = new int[A.length];
        while (i <= j) {
            if (A[i] * A[i] < A[j] * A[j]) result[k--] = A[j] * A[j--];
            else result[k--] = A[i] * A[i++];
        }
        return result;
    }
}

Python:

class Solution:
    def sortedSquares(self, A: List[int]) -> List[int]:
        return sorted((i ** 2 for i in A))