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.
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
#include <iostream>
struct ListNode
{
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
class Solution
{
public:
ListNode *partition(ListNode *head, int x)
{
ListNode *LessHeader = new ListNode(10);
ListNode *LessTail = LessHeader;
ListNode *GreaterHeader = new ListNode(10);
ListNode *GreaterTail = GreaterHeader;
while(head != NULL)
{
if (head->val < x)
{
LessTail->next = head;
LessTail = head;
}
else
{
GreaterTail->next = head;
GreaterTail = head;
}
head = head->next;
}
LessTail->next = GreaterHeader->next;
GreaterTail->next = NULL;
delete GreaterHeader;
head = LessHeader->next;
delete LessHeader;
return head;
}
};
void Test()
{
ListNode *head = new ListNode(2);
head->next = new ListNode(1);
int x = 2;
Solution S1;
ListNode *Result = S1.partition(head, x);
while (Result != NULL)
{
std::cout << "The result is " << Result->val << std::endl;
Result = Result->next;
}
}
原文地址：http://blog.csdn.net/sheng_ai/article/details/45042429
