LeetCode初级算法【链表专题】

1. 删除链表的倒数第N个节点

给定一个链表，删除链表的倒数第 n 个节点，并且返回链表的头结点

思路 链表常用解法：双指针定位

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode removeNthFromEnd(ListNode head, int n) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode pre = dummy;
        ListNode cur = dummy;
        while(n--!=0){
             cur = cur.next;
        }
        while(cur.next!=null){
             cur = cur.next;
             pre = pre.next;
        }
        pre.next = pre.next.next;
        return dummy.next;
    }
}

2. 反转链表

迭代或递归地反转链表。你能否用两种方法解决这道题？

递归+回溯解法：

	public ListNode reverseList(ListNode head) {
        if(head == null || head.next==null){
            return head;
        }
        //获得尾节点
        ListNode ret = reverseList(head.next);
        //下面属于回溯，从后往前继续切换
        head.next.next = head;
        head.next = null;
        return ret;
    }

迭代：

	public ListNode reverseList(ListNode head) {
        ListNode cur = head;
        ListNode pre = null;
        while(cur!=null){
            ListNode next = cur.next;
            cur.next = pre;
            pre = cur;
            cur = next;
        }
        return pre;
    }

递归思想还是有一定难度的，注意是进入递归前处理还是递归后处理，递归后处理(回溯)要有返回上个现场的思想，这道题需要在递归后才好处理节点转向。

3. 合并两个有序链表

思路1：循环拼接

	public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        ListNode pre = new ListNode(-1);
        ListNode prev = pre;
        while(l1!=null && l2!=null){
            if(l1.val<l2.val){
                prev.next = l1;
                l1 = l1.next;
            }else{
                prev.next = l2;
                l2 = l2.next;
            }
            prev = prev.next;
        }
        prev.next = l1==null?l2:l1;
        return pre.next;
    }

思路2：递归

    public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if(l1==null){
            return l2;
        }
        if(l2==null){
            return l1;
        }
        if(l1.val<l2.val){
            l1.next = mergeTwoLists(l1.next,l2);
            return l1;
        }else{
            l2.next = mergeTwoLists(l1,l2.next);
            return l2;
        }
    }

4. 是否为回文链表

时间复杂度O(n),空间复杂度o(2n)

	public boolean isPalindrome(ListNode head) {
        Stack<Integer> stack = new Stack<>();
        Queue<Integer> que = new LinkedList<>();
        while(head!=null){
            que.add(head.val);
            stack.add(head.val);
            head = head.next;
        }
        while(!que.isEmpty()){
            if(!que.poll().equals(stack.pop())){
                return false;
            }
        }
        return true;
    }

5. 检测链表的环入口

如果链表有环，返还入口节点，否则返回null

典型的双指针问题：假设快慢指针有交点，则x+y+z+y = 2(x+y)

	public ListNode detectCycle(ListNode head) {
        ListNode fast = head, slow = head;
        while (true) {
            if (fast == null || fast.next == null) return null;
            fast = fast.next.next;
            slow = slow.next;
            if (fast == slow) break;
        }
        fast = head;
        while (slow != fast) {
            slow = slow.next;
            fast = fast.next;
        }
        return fast;
    }