《劍指offer》面試題25：合併兩個排序的鏈表

題目描述

知識點：列表、遞歸

 

思路一：遞歸

pMergedHead指向頭節點值小的鏈表頭，遞歸對剩下兩段鏈表合併

# -*- coding:utf-8 -*-
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None
        
class Solution:
    # 返回合併後列表
    def Merge(self, pHead1, pHead2):
        # write code here
        if not pHead1:
            return pHead2
        if not pHead2:
            return pHead1 
        
        pMergedHead = None
        
        if pHead1.val <= pHead2.val:
            pMergedHead = pHead1
            pMergedHead.next = self.Merge(pHead1.next, pHead2)
        elif pHead2.val <= pHead1.val:
            pMergedHead = pHead2
            pMergedHead.next = self.Merge(pHead1, pHead2.next)
                
        return pMergedHead

思路二：非遞歸

pMergedHead和cur指向值小的鏈表頭，cur爲遊標，後面依次連接值小的節點

# -*- coding:utf-8 -*-
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def Merge2(self, pHead1, pHead2):
        # write code here
        if not pHead1:
            return pHead2
        elif not pHead2:
            return pHead1 
        
        cur = pMergedHead = None
        if pHead1.val <= pHead2.val:
            cur = pMergedHead = pHead1
            pHead1 = pHead1.next
        elif pHead2.val <= pHead1.val:
            cur = pMergedHead = pHead2
            pHead2 = pHead2.next
        
        while pHead1 and pHead2:
            if pHead1.val <= pHead2.val:
                cur.next = pHead1
                cur = cur.next
                pHead1 = pHead1.next
            elif pHead2.val <= pHead1.val:
                cur.next = pHead2
                pHead2 = pHead2.next
                cur = cur.next
        if pHead1:
            cur.next = pHead1
        elif pHead2:
            cur.next = pHead2

        return pMergedHead