问题
I read some interview question online about how would you find if there's a loop in a linked list, and solution (Floyd's cycle-finding algorithm) is to have two pointers, one is 2x faster than the other, and check if they meet again.
My question is: Why can't I just keep one pointer fixed, just move the other pointer forward by 1 step each time?
回答1:
Because the first (non-moving) pointer might not lie within the loop, so the pointers would never meet. (Remember that a loop may consist of only part of the list.)
回答2:
Because maybe not the complete linkedList is within the loop.
For a linked list lasso detection algorithm, you need two pointers:
As long as the first pointer is where the cowboy is, no loop is detected. But if you move it stepwise forward, it will eventually enter the loop.
BTW, lasso is the terminus technicus from graph theory, cowboy isn't.
回答3:
Because the loop may not contain the element pointed to by the first pointer.
For example, if the first pointer points to element 1 and the linked list has a loop later on (1->2->3->4->2), your algorithm won't detect it.
来源：https://stackoverflow.com/questions/7398703/linked-list-loop-detection-algorithm