leetcode solution: Remove Nth Node From End of List (19)

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

For example,

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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.

解法1： Two pointers, One Pass, O(N) Time

头node不确定，上dummy node大法。然后用双指针向前移动，找到Nth node的前一个node，删除后返回dummy->next
C++

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/**
 * 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) {
        ListNode* dummy = new ListNode(0);
        dummy->next = head;
        ListNode* left = dummy;
        head = dummy;
        for (int i = 0; i < n; ++i) {
            head = head->next;
        }
        
        while (head->next) {
            left = left->next;
            head = head->next;
        }
        ListNode* temp = left->next;
        left->next = left->next->next;
        ListNode* res = dummy->next;
        delete temp;
        delete dummy;
        return res;
    }
};

Java

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