algorithm - 使用 Kadane 算法查找数组中最大和子数组的数量

Question

Kadane 算法 ( http://en.wikipedia.org/wiki/Maximum_subarray_problem ) 用于查找一维数字数组中连续子数组的最大和。

现在，如何使用它来找出具有相同最大和的此类序列的数量？可以对算法进行哪些修改以计算此类序列..

例如:

0 0 0 1     -> (largest sum = 1); 4 sequences  { (0,0,0,1), (0,0,1), (0,1) , (1) }

0 0 0       -> (largest sum = 0); 6 sequences { (0,0,0), (0,0), (0,0), (0), (0), (0) }

2 0 -2 2    -> (largest sum = 2); 4 sequences { (2), (2,0), (2,0,-2, 2), (2) }

Answer 1

Kadane 的算法跟踪在当前点结束的序列的最大值，以及到目前为止看到的最大值。

这里是一个基于the wikipedia page的Python实现:

def kadane(A):
    max_ending_here = max_so_far = 0
    for x in A:
        max_ending_here = max([x,0,max_ending_here+x])
        max_so_far = max(max_so_far,max_ending_here)
    return max_so_far

我们可以修改算法，通过添加两个变量来跟踪此类序列的计数:

• count_with_max_ending_here 计算在当前点结束且值总和为 max_ending_here 的序列数
• count_with_max 统计目前找到的最大值的序列数

可以直接更改 Kadane 的算法以跟踪这些变量，同时保持 O(n) 的复杂度:

def kadane_count(A):
    max_ending_here = max_so_far = 0
    count_with_max_ending_here = 0 # Number of nontrivial sequences that sum to max_ending_here
    count_with_max = 0
    for i,x in enumerate(A):
        new_max = max_ending_here+x
        if new_max>=0:
            if count_with_max_ending_here==0:
                # Start a nontrivial sequence here
                count_with_max_ending_here=1
            elif max_ending_here==0:
                # Can choose to carry on a nontrivial sequence, or start a new one here
                count_with_max_ending_here += 1
            max_ending_here = new_max
        else:
            count_with_max_ending_here = 0 
            max_ending_here = 0

        if max_ending_here > max_so_far:
            count_with_max = count_with_max_ending_here
            max_so_far = max_ending_here
        elif max_ending_here == max_so_far:
            count_with_max += count_with_max_ending_here

    return count_with_max

for A in [ [0,0,0,1],[0,0,0],[2,0,-2,2] ]:
    print kadane(A),kadane_count(A)