一、题目

二、代码

class Solution {
    
public:
    vector<int> sortedSquares(vector<int>& nums) {
    
    
     //本题思路:First divide the array into positive and negative Array ordered by absolute value 然后再处理
     
     //second thought directly found location There is no need to declare two arrays 干就完事了
     //常规变量
     int i=0;
     int j=0;
     int k=0;
     //本题变量
     int size=0;
     size= nums.size();  
     vector<int> vector_for_return(size,0);	   //迭代器的使用,必须要先用push_back()或者={}初始化,才可以
    


         if(size!=0)
         {
    
            std::cout<<"size "<<size<<std::endl;
             int record_positive=0;

            

             // int vector_for_return[size]; //数组声明并且清零
             // for(i=0;i<=size-1;i++) vector_for_return[i]=0;


            //首先需要找到0 作为分界点
            if(nums[0]>=0)           //不存在0的情况 全部大于0
            {
    
            record_positive=-1;
            }
            else if(nums[size-1]<0)   //不存在0的情况 全部都小于0
            {
    
            record_positive=-2; 
            }
            else
            {
    
                for(i=0;i<=size-1;i++) 
                {
    
                    if(nums[i]>=0)  
                    {
    
                        record_positive=i;
                        break;
                    }
                }
            }
          
          std::cout<<"record_positive "<<record_positive<<std::endl;

             
            if(record_positive==-1)   for(i=0;i<=size-1;i++) vector_for_return[i]=nums[i]*nums[i];
            
            if(record_positive==-2)    for(i=size-1;i>=0;i--) vector_for_return[(size-1)-i]=nums[i]*nums[i];
            
             if(record_positive!=-1&&record_positive!=-2)
            {
    
               j=record_positive-1;
               k=record_positive;
               std::cout<<"j "<<j<<std::endl;
               std::cout<<"k "<<k<<std::endl;
              for(i=0;i<=size-1&&j>=0&&k<=size-1;i++)
              {
    
                  std::cout<<"j in "<<j<<std::endl;
                  std::cout<<"k in "<<k<<std::endl;
                 if(nums[j]*nums[j]<=nums[k]*nums[k])  
                 {
    
                     vector_for_return[i]=nums[j]*nums[j];
                     j=j-1;
                     std::cout<<"j final "<<j<<std::endl;
                      std::cout<<"i"<<i<<std::endl;
                     std::cout<<"vector_for_return[i]"<<vector_for_return[i]<<std::endl;
                 }
                  else
                 {
    
                     vector_for_return[i]=nums[k]*nums[k];
                     k=k+1;
                     std::cout<<"i"<<i<<std::endl;
                     std::cout<<"vector_for_return[i]"<<vector_for_return[i]<<std::endl;
                     std::cout<<"k final"<<k<<std::endl;
                 }

              }

              //The situation after the assignment is complete should be side to end The other side has not come 
             if(j==-1&&k!=size) 
             {
    
                 for(;i<=size-1,k<=size-1;i++,k++)   vector_for_return[i]=nums[k]*nums[k];
             }
              if(j!=-1&&k==size) 
             {
    
                 for(;i<=size-1,j>=0;i++,j--)   vector_for_return[i]=nums[j]*nums[j];
             }

             

             }


         }
         else ;

        //for(i=0;i<=size-1;i++) std::cout<<"i "<<i<<"vector_for_return[i] "<<vector_for_return[i]<<std::endl;
        return vector_for_return;
    }
};

三、运行结果