随机洗牌

转自：cracking the code interview

Write a method to shuffle a deck of cards. It must be a perfect shuffle - in other words, each 52! permutations of the deck has to be equally likely. Assume that you are given a random number generator which is perfect.

SOLUTION

This is a very well known interview question, and a well known algorithm. If you aren’t one of the lucky few to have already know this algorithm, read on.

Let’s start with a brute force approach: we could randomly selecting items and put them into a new array. We must make sure that we don’t pick the same item twice though by somehow marking the node as dead.

Array: [1] [2] [3] [4] [5]
Randomly select 4: [4] [?] [?] [?] [?]
Mark element as dead: [1] [2] [3] [X] [5]
The tricky part is, how do we mark [4] as dead such that we prevent that element from being picked again? One way to do it is to swap the now-dead [4] with the first element in the array:

Array: [1] [2] [3] [4] [5]
Randomly select 4: [4] [?] [?] [?] [?]
Swap dead element: [X] [2] [3] [1] [5]

Array: [X] [2] [3] [1] [5]
Randomly select 3: [4] [3] [?] [?] [?]
Swap dead element: [X] [X] [2] [1] [5]

By doing it this way, it’s much easier for the algorithm to “know” that the first k elements are dead than that the third, fourth, nineth, etc elements are dead. We can also optimize this by merging the shuffled array and the original array.

Randomly select 4: [4] [2] [3] [1] [5]
Randomly select 3: [4] [3] [2] [1] [5]

This is an easy algorithm to implement iteratively:

public static void shuffleArray(int[] cards) {
	
	int temp, index;
	
	for (int i = 0; i < cards.length; i++){
		
		index = (int) (Math.random() * (cards.length - i)) + i;		
		temp = cards[i];		
		cards[i] = cards[index];		
		cards[index] = temp;
	
	}
}