LeetCode(19) Remove Nth Node From End of List
2013-12-18 14:30
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题目如下:
Given a linked list, remove the Nth node from the end of list and return its head.
For example,
Given linked list: 1->2->3->4->5, and n = 2.
After removing the second node from the end, the linked list becomes 1->2->3->5.
Note:
Given n will always be valid.
Try to do this in one pass.
题目比较简单。注意分为删除头结点和非头结点这两种情况即可。提交的头文件如下:
我是用的main函数如下:
update: 2014-12-10
update: 2015-01-22
更巧妙的思路:
一个指针先走n步,然后两个同步走,直到第一个走到终点,第二个指针就是需要删除的节点。注意讨论head指针的情况。
Given a linked list, remove the Nth node from the end of list and return its head.
For example,
Given linked list: 1->2->3->4->5, and n = 2.
After removing the second node from the end, the linked list becomes 1->2->3->5.
Note:
Given n will always be valid.
Try to do this in one pass.
题目比较简单。注意分为删除头结点和非头结点这两种情况即可。提交的头文件如下:
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *removeNthFromEnd(ListNode *head, int n) { if(head==NULL) return NULL; int list_length=0; int index1=0; ListNode* p=head; ListNode* q=NULL; while(p!=NULL){ list_length++; p=p->next; } if(list_length<n) { //题目说了给定的n是valid的。貌似不需要此步的测试了。 return NULL; }else if(list_length==n){ p=head; head=head->next; delete p; return head; }else { index1=list_length-n+1; p=head; for(int i=0;i<index1-2;i++) p=p->next; q=p->next; p->next=p->next->next; delete q; return head; } } };
我是用的main函数如下:
// main.cpp #include <iostream> #include "Solution.h" int main(int argc, const char * argv[]) { int array[5]={-1,-2,-3,-4,-5}; ListNode* head=NULL; ListNode* tail=NULL; for(int i=0;i<5;i++) { ListNode* p=new ListNode(array[i]); if(head==NULL){ head=p; tail=p; }else{ tail->next=p; tail=p; } } ListNode* p=head; while(p!=NULL){ std::cout<<p->val<<"\t"; p=p->next; } Solution s1; head=s1.removeNthFromEnd(head, 100); std::cout<<std::endl; std::cout << "019 Hello, World!\n"; while(head!=NULL){ std::cout<<head->val<<"\t"; head=head->next; } std::cout<<std::endl; return 0; }
update: 2014-12-10
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode *removeNthFromEnd(ListNode *head, int n) { int length = 0; ListNode* current = head; while (current != NULL) { length++; current = current -> next; } n = length - n -1; // previous node of the node to be deleted current = head; if (n < 0) // NOTE: corner case, delete head return head->next; while (n > 0) { current = current->next; n--; } current->next = current->next->next; return head; } };
update: 2015-01-22
更巧妙的思路:
一个指针先走n步,然后两个同步走,直到第一个走到终点,第二个指针就是需要删除的节点。注意讨论head指针的情况。
class Solution { public: ListNode *removeNthFromEnd(ListNode *head, int n) { if (n == 0) return head; ListNode* first = head; ListNode* second = head; int step = 0; while (step < n ) { first = first->next; step++; } if (first == NULL) return head->next; while (first!= NULL && first->next != NULL) { first = first->next; second = second->next; } if(second->next != NULL) second->next = second->next->next; return head; } };
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