Leetcode 53. Maximum Subarray

標籤：

Find the contiguous subarray within an array (containing at least one number) which has the largest sum.

For example, given the array [−2,1,−3,4,−1,2,1,−5,4],
the contiguous subarray [4,−1,2,1] has the largest sum = 6.

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More practice:

If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

[Solution]dynamic programming:let local_suffix[0...i....n] represet max subarray contains the ith number in A[],

if (local_suffix[i - 1] <= 0)　　local_suffix[i] = A[i - 1];else　　local_suffix[i] = local_suffix[i - 1] + A[i - 1];

 1 int maxSubArray(int A[], int n)  2     { 3         if (n <= 0) 4             return -1; 5         int global_suffix = INT_MIN, *local_suffix = new int[n + 1]; 6          7         local_suffix[0] = 0; 8         for (int i = 1; i <= n; i++) 9         {10             if (local_suffix[i - 1] <= 0)11                 local_suffix[i] = A[i - 1];12             else13                 local_suffix[i] = local_suffix[i - 1] + A[i - 1];14             15             if (global_suffix < local_suffix[i])16                 global_suffix = local_suffix[i];17         }18         19         delete[] local_suffix;20         return global_suffix;21     }

 However, this uses O(n) memory. We can use just local_suffix instead.

 1     int maxSubArray(int A[], int n)  2     { 3         if (n <= 0) 4             return -1; 5         int global_suffix = INT_MIN, local_suffix; 6          7         for (int i = 0; i < n; i++) 8         { 9             if (local_suffix <= 0)10                 local_suffix = A[i];11             else12                 local_suffix = local_suffix + A[i];13             14             if (global_suffix < local_suffix)15                 global_suffix = local_suffix;16         }17         18         return global_suffix;19     }

 

Leetcode 53. Maximum Subarray