Leetcode 53. Maximum Subarray-优快云博客

本文介绍了一种寻找含至少一个数的连续子数组并求其最大和的算法。以[-2,1,-3,4,-1,2,1,-5,4]为例，[4,-1,2,1]具有最大和为6。文章提供了动态规划解决方案，并讨论了使用分治法的另一种实现。

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.

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             else
13                 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             else
12                 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     }