[LeetCode] 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.
问题描述:给定一个链表,从链尾开始算起,删除第n个节点,并返回链表头。
看到这题,我就想到了递归,因为不采用递归的话,就无法确定倒数第n个节点。采用递归,当遍历到链尾时就是0,最后一个节点就是1,依次类推。
class Solution { public: int listLen(ListNode *head) { ListNode *p = head; int n = 0; while(p) { p = p->next; ++n; } return n; } int removenode(ListNode *head, int n) { if(head == NULL) return 0; int m = removenode(head->next, n) + 1; if(m == n + 1) { ListNode *p = head->next; head->next = p->next; free(p); } return m; } ListNode *removeNthFromEnd(ListNode *head, int n) { // IMPORTANT: Please reset any member data you declared, as // the same Solution instance will be reused for each test case. int len = listLen(head); if(n == len) { ListNode *p = head; head = head->next; free(p); } removenode(head, n); return head; } };
在上面的解法中,主要有两种n需要区分,当删除的节点是第一个节点时,和删除的节点不是第一个节点时,这是因为,我的removenode()函数删除的是参数后面的那个节点,当要删除的节点是第一个节点时,它没有前面的节点,因此要分开处理。
补充:软件开发 , C++ ,