链表中倒数第K个结点 牛客网 程序员面试金典 C++ Python
题目描述输入一个链表，输出该链表中倒数第k个结点。C++
/* struct ListNode { int val; struct ListNode *next; ListNode(int x) : val(x), next(NULL) { } };*/ class Solution { public: //run:3ms memory:476k ListNode* FindKthToTail(ListNode* pListHead, unsigned int k) { if(NULL == pListHead) return NULL; if(0 == k) return NULL; ListNode* p = pListHead; ListNode* res = pListHead; for(unsigned int i = 0; i< k; i++) if (p) p=p->next; else return NULL; for(;p;p=p->next) res = res->next; return res; } ListNode* FindKthToTail2(ListNode* pListHead, unsigned int k) { if(NULL == pListHead) return NULL; if(0 == k) return NULL; ListNode* p = pListHead; ListNode* res = pListHead; for(unsigned int i = 0; i< k; i++) if (p) p=p->next; else return NULL; while(p) { res = res->next; p = p->next; } return res; } };Python
# class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: #run:33ms memory:5728k def FindKthToTail2(self, head, k): if k<=0 or head == None: return None p = head ret = head for i in range(k): if p:p = p.next else:return None while(p): p = p.next ret = ret.next return ret #run:22ms memory:5852k def FindKthToTail(self, head, k): if k<=0 or head == None: return None else: count = 0 p = head ret = head while p!=None: count = count + 1 if count > k: ret=ret.next p = p.next if count < k: ret = None return ret
转载于:https://www.cnblogs.com/vercont/p/10210300.html
