zoukankan      html  css  js  c++  java
  • 3*n/2时间内求出最大最小值

    #include<iostream>
    using namespace std;
    void getMaxMin(int a[], int len)
    {
        int max, min;
        if (len % 2)
        {
            max = min = a[0];
            for (int i = 1; i < len; i += 2)
            {
                if (a[i] < a[i + 1])
                {
                    if (a[i + 1]>max)
                    {
                        max = a[i + 1];
                    }
                    if (a[i] < min)
                    {
                        min = a[i];
                    }
                }
                else
                {
                    if (a[i]>max)
                    {
                        max = a[i];
                    }
                    if (a[i + 1] < min)
                    {
                        min = a[i + 1];
                    }
                }
            }
        }
        else
        {
            if (a[0]>a[1])
            {
                max = a[0];
                min = a[1];
            }
            else
            {
                max = a[1];
                min = a[0];
            }
            for (int i = 2; i < len; i += 2)
            {
                if (a[i] < a[i + 1])
                {
                    if (a[i + 1]>max)
                    {
                        max = a[i + 1];
                    }
                    if (a[i] < min)
                    {
                        min = a[i];
                    }
                }
                else
                {
                    if (a[i]>max)
                    {
                        max = a[i];
                    }
                    if (a[i + 1] < min)
                    {
                        min = a[i + 1];
                    }
                }
            }
    
        }
        cout << "max:	" << max << endl;
        cout << "min:	" << min << endl;
    }
    int main()
    {
        int a[] = {1,4,5,2,8,9,2,0,11,22,345,1,3};
        int N = sizeof a / sizeof a[0];
        getMaxMin(a, N);
    }
  • 相关阅读:
    sql事务
    连续按两次提示退出功能
    页面跳转及传值
    TextView详解
    textAppearance的属性设置
    POJ-1459 Power Network
    POJ-2112 Optimal Milking
    POJ-1149 PIGS
    AOJ-722 发红包
    HDU-3605 Escape
  • 原文地址:https://www.cnblogs.com/liuhg/p/MAXMIN.html
Copyright © 2011-2022 走看看