找到一个数组的最大元素递归 [英] Finding the maximum element of an array recursively

问题描述

考虑这个code，它计算一个数组的最大元素。

Consider this code, which computes the maximum element of an array.

#include <stdio.h>

int maximum(int ar[], int n)
{

    if (n == 1) {
        return ar[0];

    } else {
        int max = maximum(ar, n-1);
        printf("Largest element : %d\n", max);
        return 5; // return ar[n-1] > max ? ar[n-1] : max;
    }
}

int main()
{
    int array[5] = {5, 23, 28, 7, 1};
    printf("Maximum element of the array is: %d", maximum(array, 5));
    return 0;
}

为什么叫else块四次？

Why is the else block called four times?

推荐答案

函数是递归的，所以它会被多次调用。

The function is recursive, thus it will be called multiple times.

当你第一次启动，N = 5。这将需要else块（n是不是1）。
然后，您有n-1再次调用最大值（N = 4）。同样，其他块服用。

When you first start, n=5. It will take the else block (n is not 1). Then, you call maximum again with n-1 (n=4). Again, the else block is taken.

总而言之，函数被调用4次n达到1日前，于是它采取if块并返回AR [0]

All told, the function is called 4 times before n reaches 1, whereupon it takes the if block and returns ar[0].

正如其他人提及的，作为写入的功能将不会返回列表的最大值。奇怪的是，它似乎总是返回5除非列出数组大小为1，在这种情况下，它返回该元素的值。

As others have mentioned, the function as written will not return the maximum value of the list. Curiously, it seems to always return 5 unless the list array size is 1, in which case it returns the value of that element.

相反，递归方法将通常涉及分割成两半，每次列表中，然后返回每一对的最大当列表终于分成元素对。

Instead, a recursive approach would typically involve splitting the list in half each time, then returning the max of each pair when the list finally broken into pairs of elements.

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