我很困惑于解决这个时间复杂性问题.
I am confused on solving this time complexity problem.
T(n) = T(n-1)我知道在最坏的情况下T(n) = T(n-1) + T(1) + n
I know in quick-sort worst case T(n) = T(n-1) + T(1) + n
评估为(n-1) + (n-2) + (n-3) + ... + 1&这个几何序列等于O(n^2)
Which evaluates to (n-1) + (n-2) + (n-3) + ... + 1 & this geometric sequence equals O(n^2)
但是.我在stackoverflow上看到的答案是T(n) = T(n-1) + c = O(n).
However. I see answers on stackoverflow that say T(n) = T(n-1) + c = O(n).
当它也等于(n-1) + (n-2) + (n-3) + ... + 1等于O(n^2)
有人可以解释一下吗.
推荐答案T(n) = T(n-1) + c不等于(n-1) + (n-2) + (n-3) + ... + 1，因为要添加的术语是常量.基本上:
T(n) = T(n-1) + c isn't equal to (n-1) + (n-2) + (n-3) + ... + 1 because the terms being added are constants. Basically:
未添加任何内容:
T(n) = T(n-1) 0 + 0 + 0 + ... + 0 = 0 O(1)添加常量:
T(n) = T(n-1) + c c + c + c + ... + c = nc O(n)添加变量:
T(n) = T(n-1) + n 1 + 2 + 3 + ... + n = n(n+1)/2 O(n^2)