Question

我做了以下练习：给定一个int数组，在递归中返回数组中两组不同数字之间的最低绝对差值。例如：如果您有以下数组：{5,4,2}最低差异为1，因为如果将其拆分为两组：{5}，{4,2}，您将获得：Math.abs（5-6））= 1。另一个例子：如果你有以下数组：{4,3,2,1}，最低的差异是0，因为如果你把它分成两组：{4,1}，{3,2}你得到：Math.abs （5-5）= 0。这必须只是递归，你可以创建尽可能多的方法。方法签名是： public static int minDiff（int [] arr）

我得到的是：

public static int minDiff(int[] arr){
    return minDiff(arr,0,0,0,arr[0]);
}
private static int minDiff(int[] arr,int index,int groupa, int groupb, int diff){
    if(index>=arr.length || arr.length-1-index<0 || (index==arr.length-1-index) ) return diff;
    int dif1=minDiff(arr,index+1,groupa+arr[index], groupb, diff);
    int dif2=minDiff(arr,index+1,groupa, groupb+arr[index], diff);
    if(Math.abs(dif1-dif2)<diff){
        diff=Math.abs(dif1-dif2);
    }
    return diff;

}

请告知

Answer 1

看看这个片段

public static void main(String[] args)
{
    System.out.println(minDiff(new int[] { 5, 2, 4 })); // 1
    System.out.println(minDiff(new int[] { 5, 4, 2 })); // 1
    System.out.println(minDiff(new int[] { 1, 2, 7, 17, 6 })); // 1
    System.out.println(minDiff(new int[] { 4, 3, 2, 1 })); // 0

}

public static int minDiff(int[] arr)
{
    return minDiff(arr, 0, 0, 0);
}

private static int minDiff(int[] arr, int sumGroupA, int sumGroupB, int i)
{
    if (i == arr.length) // in case we are out of boundaries return the absolute value between the two groups
        return Math.abs(sumGroupA - sumGroupB);

    int r1 = minDiff(arr, sumGroupA + arr[i], sumGroupB, i + 1); // add to group 1
    int r2 = minDiff(arr, sumGroupA, sumGroupB + arr[i], i + 1); // add to group 2 instead

    return Math.min(r1, r2);
}

陷入递归 - 找到最低的差异

1 个答案: