链表分割 牛客网 程序员面试金典 C++ Python

链表分割 牛客网 程序员面试金典 C++ Python

• 题目描述
• 编写代码，以给定值x为基准将链表分割成两部分，所有小于x的结点排在大于或等于x的结点之前
• 给定一个链表的头指针 ListNode* pHead，请返回重新排列后的链表的头指针。注意：分割以后保持原来的数据顺序不变。

c++

/*
 * struct ListNode {
 *     int val;
 *         struct ListNode *next;
 *             ListNode(int x) : val(x), next(NULL) {}
 *             };*/
class Partition {
public:
//run:4ms memory:480k
     ListNode* partition(ListNode* pHead, int x){
        if(pHead==NULL) return NULL;
        ListNode *pNode = pHead;
        ListNode *small = new ListNode(0);
        ListNode *big = new ListNode(0);
        ListNode *pSmall = small;
        ListNode *pBig = big;
        while(pNode){
            if(pNode->val<x){
                pSmall->next = pNode;
                pNode = pNode->next;
                pSmall = pSmall->next;
                pSmall->next = NULL;
            }else{
                pBig->next = pNode;
                pNode = pNode->next;
                pBig = pBig->next;
                pBig->next = NULL;
            }
        }
        pSmall->next = big->next;
        free(big);
        pNode = small->next;
        free(small);
        return pNode;
    }
};

Python

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

class Partition:
#run:94ms memeory:5688k
    def partition(self, pHead, x):
        if None == pHead: return None
        pNode = pHead
        small = ListNode(0)
        big = ListNode(0)
        pSmall = small
        pBig = big
        while pNode:
            if pNode.val < x:
                pSmall.next = pNode
                pNode = pNode.next
                pSmall = pSmall.next
                pSmall.next = None
            else:
                pBig.next = pNode
                pNode = pNode.next
                pBig = pBig.next
                pBig.next = None
        pSmall.next = big.next
        return small.next