《算法零基础100讲》 最值算法

题目1：

题目链接

int findMaxConsecutiveOnes(int* nums, int numsSize)
{
    int cnt=0;
    int max=0;
    for(int i=0;imax?cnt:max;
        }
        else
        {
            cnt=0;
        }
    }
    return max;
}

题目2
题目链接

//找最大和次最大的数
int maxProduct(int* nums, int numsSize){
    int max=nums[0];
    int smax=nums[1];
    for(int i=1;i=max)
        {
            smax=max;
            max=nums[i];
        }
        else if(nums[i]=smax)
        {
            smax=nums[i];
        }
    }
    return (max-1)*(smax-1);
}

题目3

题目链接

//二分
int findMin(int* nums, int numsSize){
    int left=0;
    int right=numsSize-1;
    while(leftnums[right])
        {
            left=mid+1;
        }
        else
        {
            right=mid;
        }
    }
    return nums[left];
}
//排序
 int change(void*num1,void*num2)
{
     return *(int*)num1-*(int*)num2;
}
 int findMin(int* nums, int numsSize)
{
     qsort(nums,numsSize,sizeof(int),change);
     return nums[0];
}

题目4

题目链接

//二分
int findMin(int* nums, int numsSize){
    int left=0;
    int right=numsSize-1;
    while(leftnums[right])
        {
            left=mid+1;
        }
        else if(nums[mid] 
题目5
 
 题目链接
 
int thirdMax(int* nums, int numsSize){
    long long first=LONG_MIN;
    long long second=LONG_MIN;
    long long third=LONG_MIN;
    if(numsSize==1)
        return nums[0];
    else if(numsSize==2)
        return nums[0]>nums[1]?nums[0]:nums[1];
    else if(numsSize>2)
    {
        for(int i=0;ifirst)
            {
                third=second;
                second=first;
                first=nums[i];
            }
            else if(nums[i]second)
            {
                third=second;
                second=nums[i];
            }
            else if(nums[i]third)
            {
                third=nums[i];
            }
        }
    }
    if(third==LONG_MIN)
        return first;
    return third;
}
 
题目6
 
 题目链接
 
int maximumProduct(int* nums, int numsSize){
    int first=INT_MIN;
    int third=INT_MIN;
    int second=INT_MIN;
    int sfirst=INT_MAX;//最小的数
    int ssecond=INT_MAX;//次小的数
    for(int i=0;isfirst&&nums[i]<=ssecond)
        {
            ssecond=nums[i];
        }
        if(nums[i]>=first)
        {
            third=second;
            second=first;
            first=nums[i];
        }
        else if(nums[i]=second)
        {
            third=second;
            second=nums[i];
        }
        else if(nums[i]=third)
        {
            third=nums[i];
        }
    }
    //考虑两个负数乘以一个正数是否大于三个正数相乘
    return first*third*second>first*sfirst*ssecond?first*third*second:first*sfirst*ssecond;
}
 
刷题愉快