Swap Nodes in Pairs 解答

Question

Given a linked list, swap every two adjacent nodes and return its head.

For example,
Given 1->2->3->4, you should return the list as 2->1->4->3.

Your algorithm should use only constant space. You may not modify the values in the list, only nodes itself can be changed.

Solution

这题的核心在于dummy node和list操作。

dummy -> 1 -> 2 -> 3 -> 4

prev　　　cur   next  tmp

我们用四个指针来完成操作。

1. 预存tmp: tmp = next.next

2. 更改next: next.next = cur

3. 更改cur: cur.next = tmp

4. 更改prev: prev.next = next

5. 更新prev, cur, next:

cur = tmp

if (cur != null): next = cur.next

prev = prev.next.next

最后返回dummy.next

我们看到循环结束的条件应该是tmp为null或者tmp.next为null.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode swapPairs(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode prev = dummy, current = head, next = head.next, tmp;
        while (next != null) {
            tmp = next.next;
            next.next = current;
            current.next = tmp;
            prev.next = next;
            current = tmp;
            prev = prev.next.next;
            if (current == null) {
                break;
            } else {
                next = current.next;
            }
        }
        return dummy.next;
    }
}