递归算法空间复杂度边界分析：seamCarve函数空间复杂度疑问

Let’s break this down clearly—recursive functions’ space complexity mostly boils down to two things: the call stack depth and any extra memory allocated outside the stack. Let’s walk through each part for your code.

1. The Call Stack: The Main Space Contributor #

Your function starts at some i (I assume this is the row index of your image, so i equals the image height minus 1 initially) and recursively calls itself with i-1 until it hits i=0. That means the call stack will have exactly i+1 frames at its deepest point (since we go from i all the way down to 0).

Each stack frame holds:

• Function parameters: int i, int j, int width (all fixed-size integers)
• Local variables: int best, int k (again, fixed-size scalars)
• Hidden overhead: Return address, saved register values (also constant size)

So each individual stack frame takes O(1) constant space. Multiply that by the stack depth of O(i), and the call stack contributes O(i) total space.

2. Extra Allocated Memory: None to Account For #

Looking at your code, there’s no dynamic memory allocation (no malloc, no growing arrays, no data structures that scale with input size). All variables live directly in the stack frame, so we don’t have any extra heap space to worry about.

Final Space Complexity Boundary #

Putting it all together:

• The worst-case (and actually, the only case) space complexity is O(i), where i is the initial row index passed to seamCarve. If i represents the height of your image (let’s call that H), this simplifies to O(H) linear space.

A quick side note: I spotted a tiny typo in your loop line—you have an extra width argument in the min call. It should probably be:

best = min(best, seamCarve(i - 1, (j+k) % width, width));

But that doesn’t affect the space complexity analysis at all!

内容的提问来源于stack exchange，提问作者xsx