劍指offer：合并兩個排序的鏈表

題目描述
輸入兩個單調遞增的鏈表，輸出兩個鏈表合成后的鏈表，當然我們需要合成后的鏈表滿足單調不減規則。

# -*- coding: utf-8 -*-
# @Time         : 2019-04-24 10:17
# @Author       : Jayce Wong
# @ProjectName  : job
# @FileName     : mergeSortedLinkedList.py
# @Blog         : https://blog.51cto.com/jayce1111
# @Github       : https://github.com/SysuJayce

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    # 返回合并后列表
    def Merge(self, pHead1, pHead2):
        # 如果任何一個鏈表為空，那么返回另一個鏈表
        if not pHead1:
            return pHead2
        if not pHead2:
            return pHead1

        # 選擇兩個頭節點中值最小的那個作為新的頭節點，然后新的頭節點的下一個節點就可以遞歸下去
        if pHead1.val < pHead2.val:
            head = pHead1
            head.next = self.Merge(pHead1.next, pHead2)
        else:
            head = pHead2
            head.next = self.Merge(pHead1, pHead2.next)
        return head

    def Merge2(self, pHead1, pHead2):
        # 也可以不用遞歸的方法，使用循環的方法來歸并鏈表
        # 借助兩個輔助節點，其中begin用于定位返回鏈表的頭節點，pre用于定位返回鏈表的當前節點
        begin = pre = ListNode(0)
        while pHead1 and pHead2:
            if pHead1.val < pHead2.val:
                pre.next = pHead1
                pre = pre.next
                pHead1 = pHead1.next
            else:
                pre.next = pHead2
                pre = pre.next
                pHead2 = pHead2.next
        # 注意處理兩個鏈表長度不一致的情況
        if not pHead1:
            pre.next = pHead2
        if not pHead2:
            pre.next = pHead1
        return begin.next

def main():
    p1 = ListNode(1)
    p1.next = ListNode(3)
    p1.next.next = ListNode(5)

    p2 = ListNode(2)
    p2.next = ListNode(4)
    p2.next.next = ListNode(6)

    p = Solution().Merge(p1, p2)
    while p:
        print(p.val)
        p = p.next

if __name__ == '__main__':
    main()