Third Maximum Number

本文介绍了一种寻找数组中第三大数的简单高效算法。通过维护三个变量记录最大、次大和第三大的数,在一次遍历过程中即可完成。文章提供了具体的实现代码,并解释了初始化这些变量为LONG_MIN的重要性。

一开始想到求第k大的数的解法,其实并不需要那么复杂。
设置三个数:max1,max2,max3,分别代表第1大,第2大,第3大的数,遍历一遍就可以了。
注意:初始值都设为LONG_MIN。

int thirdMax(vector<int>& nums) {
        long max1 = LONG_MIN,max2=LONG_MIN,max3=LONG_MIN;
        for(vector<int>::iterator it= nums.begin();it != nums.end(); ++it)
        {
            int tmp = *it;
            if(tmp > max1)
            {
                max3 = max2;
                max2 = max1;
                max1 = tmp;
            }
            else if(tmp > max2 && tmp < max1)
            {
                max3 = max2;
                max2 = tmp;
            }
            else if(tmp > max3 && tmp < max2)
            {
                max3 = tmp;
            }
        }
        if(max3 == LONG_MIN || max2 == LONG_MIN)
        {
            return max1;
        }
        return max3;
    }
A piece of paper contains an array of n integers a1, a2, ..., an. Your task is to find a number that occurs the maximum number of times in this array. However, before looking for such number, you are allowed to perform not more than k following operations — choose an arbitrary element from the array and add 1 to it. In other words, you are allowed to increase some array element by 1 no more than k times (you are allowed to increase the same element of the array multiple times). Your task is to find the maximum number of occurrences of some number in the array after performing no more than k allowed operations. If there are several such numbers, your task is to find the minimum one. Input The first line contains two integers n and k (1 ≤ n ≤ 105; 0 ≤ k ≤ 109) — the number of elements in the array and the number of operations you are allowed to perform, correspondingly. The third line contains a sequence of n integers a1, a2, ..., an (|ai| ≤ 109) — the initial array. The numbers in the lines are separated by single spaces. Output In a single line print two numbers — the maximum number of occurrences of some number in the array after at most k allowed operations are performed, and the minimum number that reaches the given maximum. Separate the printed numbers by whitespaces. Examples Inputcopy Outputcopy 5 3 6 3 4 0 2 3 4 Inputcopy Outputcopy 3 4 5 5 5 3 5 Inputcopy Outputcopy 5 3 3 1 2 2 1 4 2 Note In the first sample your task is to increase the second element of the array once and increase the fifth element of the array twice. Thus, we get sequence 6, 4, 4, 0, 4, where number 4 occurs 3 times. In the second sample you don't need to perform a single operation or increase each element by one. If we do nothing, we get array 5, 5, 5, if we increase each by one, we get 6, 6, 6. In both cases the maximum number of occurrences equals 3. So we should do nothing, as number 5 is less than number 6. In the third sample we should increase the second array element once and the fifth element once. Thus, we get sequence 3, 2, 2, 2, 2, where number 2 occurs 4 times.讲解这道题
09-21
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值