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How to find the min and max of a set of numbers in C without loops?

开发者 https://www.devze.com 2023-01-17 01:36 出处:网络
Lets say I have 10 numbers (doubles) and I have to find the sma开发者_Go百科llest number and the biggest number without using loops, how would I do so?In psuedo-code, in case this is homework:

Lets say I have 10 numbers (doubles) and I have to find the sma开发者_Go百科llest number and the biggest number without using loops, how would I do so?


In psuedo-code, in case this is homework:

min = arr[0]
max = arr[0]
for n in 1..size(arr)-1:
    if arr[n] > max:
        max = arr[n]
    if arr[n] < min:
        min = arr[n]

If, for some reason, you can't use loops (and this certainly marks it as homework - no-one in their right mind would try this without a loop), just unroll the loop:

min = arr[0]
if arr[1] < min min = arr[1]
if arr[2] < min min = arr[2]
if arr[3] < min min = arr[3]
if arr[4] < min min = arr[4]
if arr[5] < min min = arr[5]
if arr[6] < min min = arr[6]
if arr[7] < min min = arr[7]
if arr[8] < min min = arr[8]
if arr[9] < min min = arr[9]
max = arr[0]
if arr[1] > max max = arr[1]
if arr[2] > max max = arr[2]
if arr[3] > max max = arr[3]
if arr[4] > max max = arr[4]
if arr[5] > max max = arr[5]
if arr[6] > max max = arr[6]
if arr[7] > max max = arr[7]
if arr[8] > max max = arr[8]
if arr[9] > max max = arr[9]

That's not too bad for ten entries but it's going to get cumbersome as the number rises.

Or a recursive solution for the more bizarrely-minded of us :-)

def findMax (arr, cur, idx):
    if idx < 0:
        return cur
    if arr[idx] > cur
        return findMax (arr, arr[idx], idx-1)
    return findMax (arr, cur, idx-1)

def findMin (arr, cur, idx):
    if idx < 0:
        return cur
    if arr[idx] < cur
        return findMin (arr, arr[idx], idx-1)
    return findMin (arr, cur, idx-1)

max = findMax (arr, arr[9], 8)
min = findMin (arr, arr[9], 8)

But I wouldn't hand that recursive solution in - if you haven't done loops yet, it's probably well beyond the level at which your class is operating.


And, since the recursive solution is a rather nifty, no-loop, solution (and you have a near-zero chance of using it and not being found out as a plagiarist), here it is:

#include <stdio.h>

static int findMin (int *arr, int cur, int idx) {
    if (idx < 0)
        return cur;
    if (arr[idx] < cur)
        return findMin (arr, arr[idx], idx-1);
    return findMin (arr, cur, idx-1);
}

static int findMax (int *arr, int cur, int idx) {
    if (idx < 0)
        return cur;
    if (arr[idx] > cur)
        return findMax (arr, arr[idx], idx-1);
    return findMax (arr, cur, idx-1);
}

int main (void) {
    int nums[] = {27,18,28,18,28,45,93,14,15,92,65,35,89};
    int min, max;
    int x = sizeof(nums) / sizeof(*nums) - 1;

    max = findMax (nums, nums[x], x-1);
    min = findMin (nums, nums[x], x-1);

    printf ("min=%d, max=%d\n", min, max);
    return 0;
}

This outputs:

min=14, max=93

as expected but don't use it on large lists since you'll probably run out of stack space.


Biggest number recursively:

double fun( double * ar, uint idx, uint size, double max )
{
    double rv;
    if( idx < size )
    {
        if( max < ar[idx] )     { rv = fun( ar, 1 + idx, size, ar[idx] ); }
        else                    { rv = fun( ar, 1 + idx, size,     max ); }
    }
    else                        { rv = max; }
    return rv;
}

Invocation:

double ray[] = { .0, -16.0, 2.0, -4.0, 8.0, -1.0, 7.0, -3.0, 5.0, -10.0 };
printf( "max %lf\n", fun( ray, 1, sizeof( ray ) / sizeof( ray[0] ), ray[0] ) );

Smallest is almost the same:

        if( ar[idx] < min )     { rv = fun( ar, 1 + idx, size, ar[idx] ); }
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