week9-leetcode #19-Remove-Nth-Node-From-End-of-List

链接：https://leetcode.com/problems/remove-nth-node-from-end-of-list/

Question

Given a linked list, remove the *n*th node from the end of list and return its head.

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.

Solution

time complecity: O(n)

space complecity:O(n)

class Solution {public:    ListNode* removeNthFromEnd(ListNode* head, int n) {      int number = 0;      ListNode* p = head;      while (p != NULL) {        number++;        p = p->next;      }      cout << "number: " << number << endl;      int count = number-n;      p = head;      if (count == 0) {        head = head->next;        delete p;      } else {        while (count >= 0 && p != NULL) {          cout << "count: " << count << endl;          if (count == 1) {            cout << "I am in " << endl;            ListNode* q = p->next;            p->next = p->next->next;            delete q;            break;          }          p = p->next;          count--;        }      }      return head;    }};

​思路：先对链表遍历一次可得到整个链表的大小，与倒数第几的数求的差值，这个差值就是正数第几个数，再遍历一次找到待删除的节点进行相应的操作。