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難度：容易

要求：

找到單鏈表倒數(shù)第n個(gè)節(jié)點(diǎn)，保證鏈表中節(jié)點(diǎn)的最少數(shù)量為n

樣例

給出鏈表** 3->2->1->5->null**和n = 2，返回倒數(shù)第二個(gè)節(jié)點(diǎn)的值1.

思路：
先翻轉(zhuǎn)，再計(jì)算

/**
     * @param head: The first node of linked list.
     * @param n: An integer.
     * @return: Nth to last node of a singly linked list. 
     */
    ListNode nthToLast(ListNode head, int n) {
        if(head == null){
            return null;
        }
        
        head = reverse(head);
        int index = 1;
        while(head.next != null){
            if(index++ == n){
                break;
            }
            head = head.next;
        }
        return head;
    }
    
    /**
     * 原地翻轉(zhuǎn)
     */
    private ListNode reverse(ListNode head){
        ListNode prev = null;
        while(head != null){
            ListNode tmp = head.next;
            head.next = prev;
            prev = head;
            head = tmp;
        }
        return prev;
    }

思路：
雙指針，第一個(gè)指針先走n個(gè)，然后第一個(gè)指針和第二個(gè)指針同時(shí)往右遍歷，當(dāng)?shù)谝粋€(gè)指針為null時(shí)，第二個(gè)指針就為倒數(shù)第n個(gè)元素

/**
     * @param head: The first node of linked list.
     * @param n: An integer.
     * @return: Nth to last node of a singly linked list. 
     */
    ListNode nthToLast(ListNode head, int n) {
        if(head == null){
            return null;
        }
        
        ListNode dummy = head;

        for(int i = 0; i < n; i++){
            if(head == null){
                return null;
            }
            head = head.next;
        }
        
        ListNode pre = dummy;
        while(head != null){
            head = head.next;
            pre = pre.next;
        }
        
        return pre;
    }