[英]Running-time complexity of two nested loops: quadratic or linear?
int Solution::diffPossible(vector<int> &A, int B) {
for (int i = 0; i < A.size(); i++) {
for (int j = i+1; j < A.size(); j++)
if ((A[j]-A[i]) == B)
return 1;
}
return 0;
}
This is the solution to a simple question where we are supposed to write a code with time complexity less than or equal to O(n). 这是一个简单问题的解决方案,在该问题中,我们应该编写时间复杂度小于或等于O(n)的代码。 I think the time complexity of this code is O(n^2) but still it got accepted.
我认为这段代码的时间复杂度为O(n ^ 2),但仍然被接受。 So, I am in doubt please tell me the right answer.
所以,我有疑问请告诉我正确的答案。
Let's analyze the worst-case scenario, ie when the condition of the if
-statement in the inner loop, (A[j]-A[i]) == B
, is never fulfilled, and therefore the statement return 1
is never executed. 让我们分析一下最坏的情况,即当的条件
if
语句来在内部循环, (A[j]-A[i]) == B
,是永远不会满足的,因此该语句return 1
从不执行。
If we denote A.size()
as n
, the comparison in the inner loop is performed n-1
times for the first iteration of the outer loop, then n-2
times for the second iteration, and so on... 如果将
A.size()
表示为n
,则对内循环的比较对于外循环的第一次迭代执行n-1
次,然后对第二次迭代进行n-2
次,依此类推...
So, the number of the comparisons performed in the inner loop for this worst-case scenario is (by calculating the sum of the resulting arithmetic progression below): 因此,在此最坏情况下,在内部循环中执行的比较次数为(通过计算以下得出的算术级数之和):
n-1 + n-2 + ... + 1 = (n-1)n/2 = (n^2 - n)/2
^ ^
|_________________|
n-1 terms
Therefore, the running-time complexity is quadratic, ie, O(n^2), and not O(n). 因此,运行时复杂度是二次的,即O(n ^ 2),而不是O(n)。
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