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问题:
My code looks like this so far:
public class ThreeSort { public static void main(String[] args) { int num1 = Integer.parseInt(args[0]); int num2 = Integer.parseInt(args[1]); int num3 = Integer.parseInt(args[2]); int x = Math.min(num1, num2); int min = Math.min(x,num3); int z = Math.max(num1, num2); int max = Math.max(z, num3); int a = 0; int mid = 0; while (mid >= min && mid <= max) { mid = a; } System.out.println(min); System.out.println(a); System.out.println(max); }
I know how to do the min and the max but I'm having troubles with the middle one. Any idea how to do that without conditional statements?
回答1:
in this case there is a simple algorithm for it:
mid = Math.max(Math.min(num1,num2), Math.min(Math.max(num1,num2),num3));
Also as the operator ^ denotes bitwise xor. so another way is:
mid=num1^num2^num3^max^min;
EXAMPLE:
public static void main(String[] args) { System.out.println(mid(70, 3, 10)); } static int mid(int a, int b, int c) { int mx = Math.max(Math.max(a, b), c); int mn = Math.min(Math.min(a, b), c); int md = a ^ b ^ c ^ mx ^ mn; return md; }
OUTPUT: 10.
Also as OldCurmudgeon said below you can calculate the mid with below formula:
int mid = num1 + num2 + num3 - min - max;
回答2:
Put them in a List and sort it...
List<Integer> ints = new LinkedList<>(); ints.add(Integer.parseInt(args[0])); ints.add(Integer.parseInt(args[1])); ints.add(Integer.parseInt(args[2])); Collections.sort(ints); // smallest -> greatest System.out.println(ints); Collections.reverse(ints); // greatest -> smallest System.out.println(ints);
回答3:
int mid = num1 + num2 + num3 - min - max;
Sorry for briefness - posted from my phone.
It must be self-evident that the middle number is the sum of the three numbers minus the max minus the min. Would also work if max == mid or max == min or even both.
回答4:
Assuming this is some kind of puzzle/homework are you allowed to use the ternary operator?
int[] ints = {3, 1, 2}; int min = ints[0] <= ints[1] && ints[0] <= ints[2] ? ints[0] : ints[1] <= ints[0] && ints[1] <= ints[2] ? ints[1] : ints[2];
回答5:
This is how I would implement three_sort without any if statements or ternary operators. You would have to adapt this to your language of choice.
def two_sort(a, b): small = int(a < b) * a + int(a >= b) * b large = int(a >= b) * a + int(a < b) * b return small, large def three_sort(a, b, c): a, b = two_sort(a, b) a, c = two_sort(a, c) b, c = two_sort(b, c) return a, b, c
for a more general solution:
from random import randint def two_sort(a, b): small = int(a < b) * a + int(a >= b) * b large = int(a >= b) * a + int(a < b) * b return small, large return li[-1] def n_sort(li): for _ in li: for i, _ in enumerate(li[:-1]): li[i], li[i+1] = two_sort(li[i], li[i+1]) return li N = 10 li = [randint(0, 1000) for _ in range(N)] print(N_Sort(N)(*li))
回答6:
public class ThreeSort { public static void main(String[] args) { // command-line input int a = Integer.parseInt(args[0]); int b= Integer.parseInt(args[1]); int c = Integer.parseInt(args[2]); // compute the order int max=Math.max(Math.max(a,b),c); int min =Math.min(Math.min(a,b ), c); int middle = a + b + c - max - min; // output in ascending order System.out.println(min+"\n" + middle+ "\n"+ max); } }
