Question

以下代码段的时间复杂度是多少？

int[][] A = new int [n][];
for (int i=0; i<n; i++) {
    if (i % 2 == 0) // i is a multiple of 2
        A[i] = new int [n];
    else
        A[i] = new int [1];
}

for (int i=0; i<A.length; i++)
    for (int j=0; j<A[i].length; j++)
        sum = sum + A[i][j];

我理解第一个循环循环n次，然后，将有n/2行长度为n的矩阵，n/2长度为1.总时间为{{ 1}}？

Answer 1

是的，复杂性将是O（n ²）。

如何吗

一半次（即n / 2次），你将遍历n个元素=（n / 2）* n = n ² / 2.
一半次（再次，n / 2次），你将只有一个要迭代的元素=（n / 2）* 1 = n / 2.
因此，整体复杂度= O（n ² / 2 + n / 2）= O（n ²）

Answer 2

Well firstly let's decide with terminology. For example, let's put that every single operation will be to equal to 1. Let's take your code (just to be consistent - we will call this method) and go line by line.

int[][] A = new int [n][];

this will be equal to 1.

for (int i=0; i<n; i++) {

Here we have loop and in worst case it will be n.

    if (i % 2 == 0) // 1
        A[i] = new int [n]; // 1
    else
        A[i] = new int [1]; // 1
}

Above operation could be counted as 1 each.

for (int i=0; i<A.length; i++)

Loop is equal to n-elements.

    for (int j=0; j<A[i].length; j++)

Inner loop is the same n.

        sum = sum + A[i][j];

Again this will be equal to 1.

Inner loops are multiplied so you are correct, but take into account that this will be exactly big-O notation O(n²).

时间复杂性审查

2 个答案: