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19. Remove Nth Node From End of List
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.
題目大意:
找到鏈表中倒數(shù)第N個元素,刪除這個元素。
代碼如下:
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: int lengthOfList(ListNode* head) { int i = 0 ; while(head != NULL) { i++; head = head->next; } return i; } ListNode* removeNthFromEnd(ListNode* head, int n) { if(head == NULL) return NULL; ListNode* p = head; int pre = lengthOfList(head) - n ; if(pre == 0) return head->next; cout << pre<<" "<<lengthOfList(head)<<endl; while(--pre) p = p->next; p->next = p->next->next; return head; } };
2016-08-12 14:02:00
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