标签:
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.
For example,
Given 1->4->3->2->5->2
and x = 3,
return 1->2->2->4->3->5
.
解题思路:对于小于x的结点重新构成一个链表,删除原链表的该值,最后连接连个链表;
#include<iostream> #include<vector> using namespace std; //Definition for singly - linked list. struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; ListNode *partition(ListNode *head, int x) { ListNode*TmpHead = head; ListNode*PreNode = head; ListNode*LessXList = new ListNode(0); ListNode*TmpLessXhead = LessXList; while (TmpHead!=NULL){ if (TmpHead->val<x) { TmpLessXhead->next = TmpHead; TmpLessXhead = TmpLessXhead->next; if (TmpHead == head) head = head->next; else PreNode->next = TmpHead->next; } else PreNode = TmpHead; TmpHead = TmpHead->next; } TmpLessXhead->next = head; return LessXList->next; }
标签:
原文地址:http://blog.csdn.net/li_chihang/article/details/43342045