[Leetcode series] maximum continuous subcolumn recursive Solution Analysis

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Step 1. Select the intermediate element of the array. The maximum sub-sequence has two possibilities: including this element/Not including.

Step 2.

Step 2.1 If the largest sub-sequence does not contain intermediate elements, perform Step 1 on the left and right sub-sequences.

Step 2.2 If the maximum sub-sequence contains, the result is very simple, that is, the maximum suffix sub-column of the left sub-column (that is, containing the last element of the left sub-column-intermediate element) add the largest prefix of the right child column to the Child column (that is, it contains the first element of the Right child column-intermediate element)

Step 3. Return the maximum value of the three columns (the maximum value of the Left subcolumn and the maximum value of the right subcolumn ).

 1 class Solution { 2 public: 3     int maxSubArray(int A[], int n) { 4         // IMPORTANT: Please reset any member data you declared, as 5         // the same Solution instance will be reused for each test case. 6         if(n==0) return 0; 7         return maxSubArrayHelperFunction(A,0,n-1); 8     } 9 10     int maxSubArrayHelperFunction(int A[], int left, int right) {11         if(right == left) return A[left];12         int middle = (left+right)/2;13         int leftans = maxSubArrayHelperFunction(A, left, middle);14         int rightans = maxSubArrayHelperFunction(A, middle+1, right);15         int leftmax = A[middle];16         int rightmax = A[middle+1];17         int temp = 0;18         for(int i=middle;i>=left;i--) {19             temp += A[i];20             if(temp > leftmax) leftmax = temp;21         }22         temp = 0;23         for(int i=middle+1;i<=right;i++) {24             temp += A[i];25             if(temp > rightmax) rightmax = temp;26         }27         return max(max(leftans, rightans),leftmax+rightmax);28     }29 };