LeetCode105|有序数组的平方

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。

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

public class SortedSquaresTest {
    public static void main(String[] args) {
        int[] A = {-4, -1, 0, 3, 10};
        int[] sortedSquares = sortedSquares(A);
        for (int num : sortedSquares
        ) {
            System.out.print(num + "\t");
        }
    }

    public static int[] sortedSquares(int[] A) {
        if (A == null || A.length == 0) {
            return new int[0];
        }
        List<Integer> list = new ArrayList<>(A.length);
        for (int num : A) {
            list.add(Math.abs(num) * Math.abs(num));
        }
        Collections.sort(list);
        int[] result = new int[list.size()];
        for (int i = 0; i < list.size(); i++) {
            result[i] = list.get(i);
        }
        return result;
    }
}

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

public class SortedSquaresTest {
    public static void main(String[] args) {
        int[] A = {-4, -1, 0, 3, 10};
        int[] sortedSquares = sortedSquares2(A);
        for (int num : sortedSquares
        ) {
            System.out.print(num + "\t");
        }
    }

     

    public static int[] sortedSquares2(int[] A) {
        if (A == null || A.length == 0) {
            return new int[0];
        }
        int i = 0;
        int j = A.length - 1;
        int index = A.length - 1;
        int[] result = new int[A.length];
        while (index >= 0) {
            if (Math.abs(A[i]) < Math.abs(A[j])) {
                result[index--] = Math.abs(A[j]) * Math.abs(A[j]);
                j--;
            } else {
                result[index--] = Math.abs(A[i]) * Math.abs(A[i]);
                i++;
            }
        }
        return result;
    }
}