lintcode 关于链表快慢指针

倒数第k个节点：O(N)时间复杂度O(1)空间复杂度

假设当前链表的长度m,要求倒数第k个链表结点

①首先建立两个指针引用，分别为slow,fast。一开始都为head头结点

②fast指针移动k-1次，此时fast指针位于链表的第k个结点位置处。

③fast与slow指针同时向后移动，当fast指针指向链表末尾时，slow指针位于链表倒数第k个位置处。

lintcode 166

class Solution:
    """
    @param: head: The first node of linked list.
    @param: n: An integer
    @return: Nth to last node of a singly linked list. 
    """
    
    def nthToLast(self, head, n):
        # write your code here
        if head is None or n < 1:
            return None
        length = head
        num = 1
        while length.next:
            length = length.next
            num += 1 
        if n > num:
            return None
        slow = head
        for i in range(n):
            head = head.next
        while head:
            head = head.next
            slow = slow.next
        return slow

链表判环：

快慢指针起点相同,快指针移动速度是慢指针两倍,当快指针与慢指针相遇的时候说明此链表有环,否则没环.

lintcode 102

class Solution:
    """
    @param head: The first node of linked list.
    @return: True if it has a cycle, or false
    """
    def hasCycle(self, head):
        # write your code here
        if head is None:
            return False
        slow,fast = head,head
        while True:
            if fast.next:
                slow = slow.next
                fast = fast.next.next
                if slow is None or fast is None:
                    return False
                elif slow == fast:
                    return True
                
            else:
                return False
        return False

找环入口：

①fast指针先行每次移动两步，slow指针后行每次移动一步，直到fast指针追上slow指针，进入②；

②slow指针从头开始往下每次移动一步，同时从第一个fast指针每次也移动一步，直到slow指针追上fast指针，该节点就是环的入口节点。

lintcode 103

class Solution:
    """
    @param head: The first node of linked list.
    @return: The node where the cycle begins. if there is no cycle, return null
    """
    def detectCycle(self, head):
        # write your code here
        if head is None or head.next is None:
            return None
        slow,fast = head, head
        dummy = head.next.next
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
            if fast == slow:
                slow = head
                while slow != fast:
                    slow = slow.next
                    fast = fast.next
                return slow

链表重排：

lintcode 99

通过快慢指针，找到链表的中点，从中点的下一个节点开始，逆序后面的链表，注意需要在链表断开处将中点节点的next指针置NULL。逆序之后，在顺序插入到前面的链表中。终止条件：两个链表中一个到达末尾。