一、题目

二、代码

class Solution {
    
public:

    int minSubArrayLen(int target, vector<int>& nums)
     {
    
         //方法3:划窗
         //我的理解是:swipe window shorter,The front is meaningless

         //常规变量
         int i=0;
         int j=0;
         int k=0;

        //本题变量
        int first_subscript=0;   //头指针
        int final_subscript=0;   //尾指针
        
        int return_length=INT32_MAX;
        int size=nums.size();
        double sum=0;
         //尾指针不断后移
        for(first_subscript=0;first_subscript<=size-1;first_subscript++)
        {
    
            sum=sum+nums[first_subscript];
            while(sum>=target)
            {
    
              double temp_length=first_subscript-final_subscript+1;   //After Satisfaction and Greater Than 比较长度
              if(temp_length<return_length) return_length=temp_length;


               sum=sum-nums[final_subscript];
               final_subscript=final_subscript+1;                   //Try shortening the length
            }
        }

        //if the length has not been modified 就赋值为0
        if(return_length==INT32_MAX) return_length=0;
        return return_length;

        // //常规变量
        // int i=0;
        // int j=0;
        // int k=0;
        // //本题变量
        // int size=nums.size();
        // int minsize=0;
        // int maxsize=0;
        // int nowsize=0;
        // int return_size=0;
         
        // //First determine whether the upper limit is met The upper limit exists to be looked up 否则返回0
        // double sum=0;
        // double temp=0;



        //方法一
         //Traversing to find this length has a timeout issue 
        //Now the time complexity is N方 Binary search can becomeN*log2n A dichotomy should not be about finding a point Instead, find an interval
        // for(i=1;i<=size;i++) //Find the length in turn1 2 3 4 .. and the largest array sumtarget比较
        // {
    
        // double sum=0;
        // for(j=0;j<=(size)-i;j++)
        // {
    
        // double temp=0;
        // for(k=j;k<=j+i-1;k++)
        // temp=temp+nums[k];
        // if(sum<temp) sum=temp;
        // }
        // if(sum>=target)
        // {
    
        // return_num=i;
        // break;
        // }
        // }

         //方法二
         //第一次改进:Use binary search for length maxsize 和 minsize Continue to reduce the length of the interval 直到max和min重合 仍然超时 无语子


        // for(i=0;i<=size-1;i++) temp+=nums[i]; 
        // if(temp<target) //If the sum of all is less than the target value 返回值为0 
        // {
    
        // return_size=0;
        // }
        // else //If the sum of all is greater than the target value 开始查找
        // {
    
        // maxsize=size;
        // while(maxsize-minsize>=2)
        // {
    
        // nowsize=(maxsize+minsize)/2;
               
        // sum=0;
        // for(j=0;j<=(size)-nowsize;j++)
        // {
    
        // temp=0;
        // for(k=j;k<=j+nowsize-1;k++)
        // temp=temp+nums[k];
        // if(sum<temp) sum=temp;
        // }
        // if(sum>=target) //如果和比目标值大 The median value is ok It can even be reduced down
        // {
    
        // maxsize=nowsize;

        // }
        // if(sum<target) //If the sum is compared to the target value The median value is not allowed must be added
        // {
    
        // minsize=nowsize+1;
        // }
        // }

        // //跳出循环时,第一种情况 max和min相等 第二种情况 min比max小 maxYes, but because of the problem of binary values, it cannot be reduced
        // if(minsize==maxsize) return_size=minsize; //相等直接返回
        // if(minsize!=maxsize) //It's not equal to see if it's okay to be small Small can be returned Otherwise return large
        // {
    
        // sum=0;
        // for(j=0;j<=(size)-minsize;j++)
        // {
    
        // temp=0;
        // for(k=j;k<=j+minsize-1;k++)
        // temp=temp+nums[k];
        // if(sum<temp) sum=temp;
        // }

        // if(sum>=target) return_size=minsize; 
        // else return_size=maxsize; 
        // }
        // }


    }


};

三、运行结果