LintCode:Swap Two Nodes in Linked List

LintCode:Swap Two Nodes in Linked List

# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None

class Solution:
    # @param {ListNode} head, a ListNode
    # @oaram {int} v1 an integer
    # @param {int} v2 an integer
    # @return {ListNode} a new head of singly-linked list
    def swapNodes(self, head, v1, v2):
        # Write your code here
        if head == None or v1 == v2:
            return head
        p1 = head
        p2 = head

        if head.val != v1 and head.val != v2:
            while p1.next != None and p1.next.val != v1:
                p1 = p1.next
            while p2.next != None and p2.next.val != v2:
                p2 = p2.next

            if p1.next == None or p2.next == None:
                return head

            if p1.next != p2 and p2.next != p1: 

                p3 = p2.next
                p5 = p2.next.next
                p3.next = p1.next.next

                p4 = p1.next
                p4.next = p5

                p1.next = p3
                p2.next = p4

                return head
            elif p2.next == p1:
                p3 = p1.next
                p4 = p1.next.next

                p3.next = p1
                p1.next = p4
                p2.next = p3
                return head

            elif p1.next == p2:
                p3 = p2.next
                p4 = p2.next.next

                p3.next = p2
                p2.next = p4
                p1.next = p3
                return head

        if head.val == v2:
            while p1.next != None and p1.next.val != v1:
                p1 = p1.next
            if p1.next == None or p2.next == None:
                return head

            if p1 != p2:
                p3 = p1.next
                p4 = p1.next.next
                p5 = p2.next

                p3.next = p5
                p1.next = p2
                p2.next = p4

                head = p3
                return head

            if p1 == p2:
                p3 = p1.next
                p4 = p1.next.next
                p5 = p2.next

                p2.next = p4
                p3.next = p2

                head = p3
                return head


        if head.val == v1:
            while p2.next != None and p2.next.val != v2:
                p2 = p2.next
            if p1.next == None or p2.next == None:
                return head