My code:
/**
* 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(Integer.MIN_VALUE);
dummy.next = head;
ListNode pre = dummy;
ListNode currL = pre.next;
ListNode currR = currL.next;
while (pre != null) {
if (pre.next == null)
break;
else
currL = pre.next;
if (currL.next == null)
break;
else
currR = currL.next;
ListNode temp = currR.next;
pre.next = currR;
currR.next = currL;
currL.next = temp;
pre = currL;
}
return dummy.next;
}
}
My test result:
Paste_Image.png
这道题目没什么好说的，虽说是medium，但是挺简单。感觉写过insertion sort list 之后，链表类的简单题，中等题都不虚了。。。
感觉代码写的还不是很好。让人感觉太多判断，不顺其自然。
改写了下，如下：
public ListNode swapPairs(ListNode head) {
if (head == null || head.next == null)
return head;
ListNode dummy = new ListNode(Integer.MIN_VALUE);
dummy.next = head;
ListNode pre = dummy;
ListNode slow = pre.next;
ListNode fast = slow.next;
while (fast != null) {
ListNode temp = fast.next;
pre.next = fast;
fast.next = slow;
slow.next = temp;
pre = slow;
slow = pre.next;
if (slow == null)
break;
fast = slow.next;
}
return dummy.next;
}
**
总结： Linkedlist, swap nodes
**
Anyway, Good luck, Richardo!
其实就是
- Reverse Nodes in k-Group -
//www.greatytc.com/p/25f5269c729d
的特殊情况。
思路很简单，没什么好说的。
My code:
/**
* 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 tail = head;
ListNode pre = dummy;
int counter = 0;
while (tail != null) {
counter++;
if (counter >= 2) {
ListNode temp = tail.next;
/** swap these two nodes */
tail.next = head;
head.next = temp;
pre.next = tail;
pre = head;
head = temp;
tail = temp;
counter = 0;
}
else
tail = tail.next;
}
return dummy.next;
}
}
Anyway, Good luck, Richardo!
My code:
/**
* 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 p1 = dummy.next;
ListNode p2 = p1;
ListNode pre = dummy;
while (p1 != null && p1.next != null) {
p2 = p1.next;
ListNode temp = p2.next;
pre.next = p2;
p2.next = p1;
p1.next = temp;
pre = p1;
p1 = temp;
p2 = temp;
}
return dummy.next;
}
}
差不多的解法。
Anyway, Good luck, Richardo! -- 08/15/2016
