SOLUTION 1: 1.将链表分逐个遍历,按大小分别加入ListL和ListR; 2.将ListL和ListR连接起来; 注意:1.ListL和ListR加入dummyNode可减少判断次数; 2.连接时ListL为空的情况; 3.ListR的tail->next要置为NULL; View Code
2. After 1 done, just link the two links. View Code CODE: https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/list/Partition.java
“穿针引线”法。 构造两个dummy,然后穿针引线。 class Solution(object): def partition(self, head, x): """ :type head: ListNode :type x: int :rtype: ListNode """ dummy1, dummy2 = ListNode(0), ListNode(0) p1, p2 = dummy1,dummy2 p1.next, p2.next = None, None i = head while...
Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x. You should preserve the original relative order of the nodes in each of the two partitions. Example: AI检测代码解析 Input: head = 1->4->3->2->5->2, x...
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题目概要 将链表中小于某个值的结点移到链表最前方。 原题链接 Partition List 题目解析 简单的数据结构题,解题思路如下: 迭代链表,将之分为两部分,一条小于x,...
博客:noahsnail.com|CSDN|简书 1. Description Partition List 2. Solution /** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */class Solution{public:ListNode*partition(ListNode*head,intx){if(!he...
Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x. You should preserve the original relative order of the...【Leetcode】86. Partition List 题目: Given a linked list and a value x, partition it such that al...
cur->next = cur->next->next; p->next =NULL; }else{ cur= cur->next; } } p->next = dummy->next;returnnewDummy->next; } }; 本文转自博客园Grandyang的博客,原文链接:划分链表[LeetCode] Partition List,如需转载请自行联系原博主。
partition_list.h #include<iostream>#include<assert.h>#include<stdlib.h>usingnamespacestd;typedefstructListNode{int_val;ListNode*_next;ListNode(intx):_val(x),_next(NULL){}}node,*node_p;classSolution{public:node_ppartition(node_p&list,intx){//参数检查if(list==NULL)returnNULL;nodedummy(-1...