求链表的中间结点

思路

一个走两步，一个走一步。一个走到尾，另外一个就走到了中间

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     struct ListNode *next;* };*/struct ListNode* middleNode(struct ListNode* head){struct ListNode* fast=head;struct ListNode* slow=head;while(1){if(head==NULL){return NULL;}fast=fast->next;if(fast==NULL){break;}slow=slow->next;fast=fast->next;if(fast == NULL){break;}}return slow;
}

求链表中倒数第k个结点

思路

让一个人先走k步，然后一起走，等最后一个为空时，前面那个就是倒数第k个结点

/*
struct ListNode {int val;struct ListNode *next;ListNode(int x) :val(x), next(NULL) {}
};*/
class Solution {
public:ListNode* FindKthToTail(ListNode* pListHead, unsigned int k) {ListNode *front = pListHead;ListNode *back = pListHead; for(int i = 0; i < k; i++ ){if(front == NULL){return NULL;}front = front->next; }while(front != NULL){front = front->next;back = back->next;}return back;}
};