[LeetCode] 019. Remove Nth Node From End of List (Easy) (C++/Python)

索引：[LeetCode] Leetcode 题解索引 (C++/Java/Python/Sql)

Github: https://github.com/illuz/leetcode

019.Remove_Nth_Node_From_End_of_List (Easy)

链接：

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

代码(github)：https://github.com/illuz/leetcode

题意：

删除一个单向链表的倒数第 N 个节点。

分析：

1. 直接模拟，先算出节点数，再找到节点删除
2. 用两个指针，一个先走 N 步，然后再一起走。

这里用 C++ 实现第一种， 用 Python 实现第二种。

Java 的话和 C++/Python 差不多，不写出来了。

代码：

C++:

class Solution {
public:
    ListNode *removeNthFromEnd(ListNode *head, int n) {
		if (n == 0)
			return head;
		// count the node number
		int num = 0;
		ListNode *cur = head;
		while (cur != NULL) {
			cur = cur->next;
			num++;
		}
		if (num == n) {
			// remove first node
			ListNode *ret = head->next;
			delete head;
			return ret;
		} else {
			// remove (cnt-n)th node
			int m = num - n - 1;
			cur = head;
			while (m--)
				cur = cur->next;
			ListNode *rem = cur->next;
			cur->next = cur->next->next;
			delete rem;
			return head;
		}
    }
};

Python:

class Solution:
    # @return a ListNode
    def removeNthFromEnd(self, head, n):
        dummy = ListNode(0)
        dummy.next = head
        p, q = dummy, dummy

        # first 'q' go n step
        for i in range(n):
            q = q.next

        # q & p
        while q.next:
            p = p.next
            q = q.next

        rec = p.next
        p.next = rec.next
        del rec
        return dummy.next