The first day of June training - array

Preface

The content of today's algorithm is ： Array

One 、 The sum of all odd length subarrays

        Here's an array of positive integers arr , Please calculate the sum of all possible odd length subarrays .
Subarray Defined as a continuous subsequence in the original array .
Please return arr in The sum of all odd length subarrays

One 、 Ideas :

         Traversal and accumulation of odd length arrays

Two 、 Source code

class Solution {
    
public:
    int sumOddLengthSubarrays(vector<int>& arr) {
    
        // Two pointers   A total number 
        int i,j,sum=0;
        for(i=0 ;i<arr.size() ;i++){
    
            // Local count   Variable 
            int count=0;
            // Local sum 
            int s=0;
          for(j=i ;j<arr.size() ;j++){
    
              // Count 
              ++count;
              // Add up 
               s+=arr[j];// error found ————> Error for ————> The subscript is wrong —— Such as ——>s+=arr[i];——————————> terms of settlement ————————> Be careful next time 
              // take   Count —————— by ————————>  An array of odd lengths ;
              if(count&1){
    
                  sum+=s;
              }
          }  
        }
        return sum;
    }
};

3、 ... and . Knowledge point

         enumeration

Two 、 Minimum distance to target element

         Give you an array of integers nums （ Subscript from 0 Start Count ） And two integers target and start , Please find a subscript i , Satisfy nums[i] == target And abs(i - start) To minimize the . Be careful ：abs(x) Express x The absolute value of .
return abs(i - start) .
Topic data assurance target Exist in nums in .

One 、 Ideas :

         Simulate directly according to the question

Two 、 Source code

class Solution {
    
public:
    int getMinDistance(vector<int>& nums, int target, int start) {
    
        // Loop variable   Minimum distance variable 
        int i,min=1000000;
        for(i=0 ;i<nums.size() ;i++){
    
            //  If the current value   be equal to   The target   And the current subscript   subtract   Given value   Is the minimum   Just assign 
            if( nums[i] == target && abs(i - start) < min){
    
                min = abs(i - start); 
            }
        }
        return min; //return 0; Return variable   The variable is written as  0 了 ...pdf
    }
};

3、 ... and . Knowledge point

         enumeration

3、 ... and 、 Dismantle the bomb

         You have a bomb to dismantle , Pressed for time ！ Your agent will give you a length of n Of loop Array code And a key k .
To get the right password , You need to replace every number . All the numbers will meanwhile Be replaced .
If k > 0 , Will be the first i For numbers Next k The sum of the numbers replaces .
If k < 0 , Will be the first i For numbers Before k The sum of the numbers replaces .
If k == 0 , Will be the first i For numbers 0 Replace .
because code It's cyclical , code[n-1] The next element is code[0] , And code[0] The first element is code[n-1] .
Here you are. loop Array code And integer keys k , Please return the decrypted results to dismantle the bomb ！

One 、 Ideas :

         Simulate by question , Module a range , Take value before or after counting

Two 、 Source code

class Solution {
    
public:
    vector<int> decrypt(vector<int>& code, int k) {
    
        int i,j,copy,val;
        int len=code.size();    

        vector<int>ret;
        for(i=0 ;i<len ;i++){
    
            val=0;
            // It can be executed 
            if(k>0){
    
                copy=i;
                for(j=0 ;j<k ;j++){
    
                    ++copy;
                    val+=code[copy%len];
                }
            }
            if(k<0){
    
                // printf("%d\n",k);
                copy=i;
                // There's a problem ,// The problem  j  And negative   End directly 
                for(j=0 ;j<abs(k) ;j++){
    
                    printf("%d\n",k);
                    --copy;
                    val+=code[(copy+len)%len];// The problem ——>copy+len-1————>val+=code[(copy+len-1)%len];———— Ben would have walked straight from the back -1 Will take one more step 
                }
            }
            // It can be executed 
            if(k==0){
    
                val=0;
            }
            ret.push_back(val);
        }
        return ret;
    }
};

// What I learned ：
// Yes vector Application （ At first, I didn't quite understand   Look at other code to know   了   operation ）2. Front and rear   Traverse trend   Go back and forth 

3、 ... and . Knowledge point

        1. Yes vector Application （ At first, I didn't quite understand Look at other code to know 了 operation ） The number to be put in Operation comes out ,ret.push_back( Target number ) that will do
         2. Front and rear Traverse trend Go back and forth （ modulus ）