寻找最接近 0 的数字

Question

我有一个整数数组，我需要找到最接近零的整数（正整数优先于负整数。）

这是我到目前为止的代码：

 public class CloseToZero {
    public static void main(String[] args) {
        int[] data = {2,3,-2};
        int curr = 0;
        int near = data[0];
        // find the element nearest to zero
        for ( int i=0; i < data.length; i++ ){
            curr = data[i] * data[i];
            if ( curr <= (near * near) )  {
                near = data[i];
            }
        }
        System.out.println( near );
    }
}

目前我得到-2的结果，但我应该得到2。我做错了什么？

原文由 user3172930 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 1

对数组进行排序（添加一行代码），如果为具有相同距离的正数和负数选择相同的绝对值，则您选择的最后一个数字将为正数。

源代码：

 import java.util.Arrays;

public class CloseToZero {

    public static void main(String[] args) {

        int[] data = {2,3,-2};
        int curr = 0;
        int near = data[0];
        Arrays.sort(data);      //  add this
        System.out.println(Arrays.toString(data));
        // find the element nearest to zero
        for ( int i=0; i < data.length; i++ ){
            System.out.println("dist from " + data[i] + " = " + Math.abs(0 -data[i]));
            curr = data[i] * data[i];
            if ( curr <= (near * near) )  {
                near = data[i];
            }
        }
        System.out.println( near );
    }
}

原文由 JuanZe 发布，翻译遵循 CC BY-SA 4.0 许可协议

Answer 2

这将在 O(n) 时间完成：

 int[] arr = {1,4,5,6,7,-1};

int closestIndex = 0;
int diff = Integer.MAX_VALUE;
for (int i = 0; i < arr.length; ++i) {
    int abs = Math.abs(arr[i]);
    if (abs < diff) {
        closestIndex = i;
        diff = abs;
    } else if (abs == diff && arr[i] > 0 && arr[closestIndex] < 0) {
        //same distance to zero but positive
        closestIndex =i;
    }
}
System.out.println(arr[closestIndex ]);

原文由 kmera 发布，翻译遵循 CC BY-SA 3.0 许可协议