求Binary Sum递归方法的时间复杂度（Big O）分析

Yep, you’ve got it exactly right—your Binary Sum method does have a time complexity of O(N²). Let’s break down why your reasoning holds up:

• First, let’s anchor on Big O basics: we only care about the dominant term as the input size ( N ) grows, and we ignore constant coefficients and lower-order terms. Your calculation estimates both recursive calls and addition operations to be roughly ( \frac{N(N+1)}{4} ). When expanded, that’s ( \frac{1}{4}N² + \frac{1}{4}N )—the ( N² ) term is the one that dictates how the runtime scales as ( N ) gets large.
• Even if your count of operations is slightly off (say, it’s actually ( \frac{N(N-1)}{2} ) instead), the key point remains: the total number of operations grows quadratically with ( N ). Constants like 1/4 or 1/2 don’t matter for Big O—we’re describing the rate of growth, not the exact number of steps.
• As a quick sanity check: if ( N = 100 ), your estimated total operations are around 2525. If ( N = 1000 ), that jumps to ~250250—notice how doubling ( N ) leads to roughly four times as many operations? That’s the hallmark of quadratic time complexity.

内容的提问来源于stack exchange，提问作者Mitch