1901. 有序数组的平方

1901. 有序数组的平方

 

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。

样例

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]

注意事项

1. 1 <= A.length <= 10000
2. -10000 <= A[i] <= 10000
3. A 已按非递减顺序排序。

public class Solution {

    /**

     * @param A: The array A.

     * @return: The array of the squares.

     */

      public int[] SquareArray(int[] A) {

            for (int i = 0; i < A.length; i++) {

                A[i] *= A[i];

            }

            mergeSort(A,0,A.length-1);

            return A;

        }

        public void mergeSort(int[] arr, int start, int end) {

            //判断拆分的不为最小单位

            if (end - start > 0) {

                //再一次拆分，知道拆成一个一个的数据

                mergeSort(arr, start, (start + end) / 2);

                mergeSort(arr, (start + end) / 2 + 1, end);

                //记录开始/结束位置

                int left = start;

                int right = (start + end) / 2 + 1;

                //记录每个小单位的排序结果

                int index = 0;

                int[] result = new int[end - start + 1];

                //如果查分后的两块数据，都还存在

                while (left <= (start + end) / 2 && right <= end) {

                    //比较两块数据的大小，然后赋值，并且移动下标

                    if (arr[left] <= arr[right]) {

                        result[index] = arr[left];

                        left++;

                    } else {

                        result[index] = arr[right];

                        right++;

                    }

                    //移动单位记录的下标

                    index++;

                }

                //当某一块数据不存在了时

                while (left <= (start + end) / 2 || right <= end) {

                    //直接赋值到记录下标

                    if (left <= (start + end) / 2) {

                        result[index] = arr[left];

                        left++;

                    } else {

                        result[index] = arr[right];

                        right++;

                    }

                    index++;

                }

                //最后将新的数据赋值给原来的列表，并且是对应分块后的下标。

                for (int i = start; i <= end; i++) {

                    arr[i] = result[i - start];

                }

            }

        }

}