編程之美 -- 尋找數組中的最大值和最小值

對於一個由N個整數組成的數組，同時找到最大值和最小值，且需要多少次比較才能着出來？

看起來好像很簡單，簡單一遍循環可以搞掂，但是效率呢？

下面是書上給出的三種方法，我用C++實現了下。。。

參數：數組，數組起始下標，數組末尾元素下標

返回：一個結構，包括最大值，最小值，比較的次數

#include <iostream>

using namespace std;

static int times = 0;

typedef struct MM {
    int max;
    int min;
    int times;
} MM;

//一個由N個整數組成的數組，同時找出最大和最小的值

//方法1：直接掃描一遍數組, 需要比較2*N次
MM findMaxAndMin_one(int *a, int begin, int end)
{
    MM mm;
    mm.max = a[begin];
    mm.min = a[begin];
    mm.times = 0;
    while(++begin <= end)
    {
        mm.times++;
        if(a[begin] > mm.max)
        {
            mm.max = a[begin];
        }
        mm.times++;
        if(a[begin] < mm.min)
        {
            mm.min = a[begin];
        }
    }
    return mm;
}

//方法2：每兩兩先比較，較大的再與當前max比較，較小的與當前min比較，比較次數：1.5*N
MM findMaxAndMin_two(int *a, int begin, int end)
{
    MM mm;
    mm.times = 0;
    int step = begin;
    mm.times++;
    if(a[step] > a[step+1])
    {
        mm.max = a[step];
        mm.min = a[step+1];
    } else
    {
        mm.max = a[step+1];
        mm.min = a[step];
    }
    step += 2;

    while(step+1 <= end)
    {
        mm.times++;
        if(a[step] > a[step+1])
        {
            mm.times++;
            if(a[step] > mm.max)
                mm.max = a[step];
            mm.times++;
            if(a[step+1] < mm.min)
                mm.min = a[step+1];
        } else
        {
            mm.times++;
            if(a[step+1] > mm.max)
                mm.max = a[step+1];
            mm.times++;
            if(a[step] < mm.min)
                mm.min = a[step];
        }
        step += 2;
    }

    //如果數組個數是奇數，則要處理最後一個值
    if(step <= end)
    {
        mm.times++;
        if(a[step] > mm.max)
            mm.max = a[step];
        mm.times++;
        if(a[step] < mm.min)
            mm.min = a[step];
    }

    return mm;
}

//方法三：分治法，只需要分別求出前後N/2個數的min和max，然後取較小的min和較大的max，遞歸下去
MM findMaxAndMin_three(int *a, int begin, int end)
{
    MM mm;
    if(end - begin <= 1)
    {
        times++;
        if(a[begin] > a[end])
        {
            mm.max = a[begin];
            mm.min = a[end];
        } else
        {
            mm.max = a[end];
            mm.min = a[begin];
        }
        return mm;
    }

    MM left_mm = findMaxAndMin_three(a, begin, begin+(end-begin)/2);
    MM right_mm = findMaxAndMin_three(a, begin+(end-begin)/2+1, end);

    times++;
    if(left_mm.max > right_mm.max)
        mm.max = left_mm.max;
    else
        mm.max = right_mm.max;

    times++;
    if(left_mm.min < right_mm.min)
        mm.min = left_mm.min;
    else
        mm.min = right_mm.min;
    return mm;
}

int main()
{
    const int LEN = 11;
    int a[LEN] = { 100, 7, 3, 5, 2, 8, 1, 4, 6, 900, 1000 };
    MM mm_one = findMaxAndMin_one(a, 0, LEN-1);
    cout << "max : " << mm_one.max << " min : " <<
    		mm_one.min << " times : " << mm_one.times << endl;

    MM mm_two = findMaxAndMin_two(a, 0, LEN-1);
    cout << "max : " << mm_two.max << " min : " <<
    		mm_two.min << " times : " << mm_two.times << endl;

    MM mm_three = findMaxAndMin_three(a, 0, LEN-1);
    mm_three.times = times;
    cout << "max : " << mm_three.max << " min : " <<
    		mm_three.min << " times : " << mm_three.times << endl;

    return 0;
}
/*
max : 1000 min : 1 times : 20
max : 1000 min : 1 times : 15
max : 1000 min : 1 times : 19
 */