A linked list and a random function are provided to design an algorithm that can return a node of the linked list at random. Each node must return the same probability. The restriction is that you can only link a table once and cannot use extra space.

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To be honest, this type of question is hard to solve if you do not know it. The possible way to think is: (assume there are n nodes in total, n is unknown) when I go to the x node, the information that can be used is that I have already taken step X. You need to decide whether to select the node or not, and the probability of the node being selected is 1/N.

 

The solution is to determine whether to select step X based on the probability of 1/X when the step X is reached. In step x + 1, the probability of step 1/(x + 1) determines whether to select step x + 1. In this way, the probability that the X is treated as the final result is that the X is selected, and the x + 1, x + 2 ,..., n) are not selected. The probability is 1/x * x/(x + 1) * (x + 1)/(x + 2)... (n-1)/n = 1/N.

ExampleCodeAs follows (input error check is not considered ):

Node * getrandomnode (node * Header)

{

Node * P = header;

Node * result = nil;

Int COUNT = 0;

For (; P! = NULL; P = p-> next)

{

Count ++;

If (0 = random (count ))

{

Result = P;

}

P = p-> next;

}

Return result;

}