描述
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.
partition_list.h
#include <iostream>
#include <assert.h>
#include <stdlib.h>
using namespace std;
typedef struct ListNode {
int _val;
ListNode *_next;
ListNode(int x) : _val(x), _next(NULL)
{}
}node,*node_p;
class Solution
{
public:
node_p partition(node_p &list,int x)
{
//参数检查
if(list==NULL)
return NULL;
node dummy(-1);
node_p head=&dummy;
head->_next=list;
node_p first=head;
//找_val为x的结点
node_p node_x=list;
while(node_x!=NULL && node_x->_val!=x){
node_x=node_x->_next;
first=first->_next;
}
if(node_x==NULL){
cout<<"not find node_x!"<<endl;
return list;
}
node_p prev=node_x;
node_p tmp=node_x->_next;
while(tmp){
if(tmp->_val<x){
prev->_next=tmp->_next;
tmp->_next=node_x;
first->_next=tmp;
first=first->_next;
tmp=node_x->_next;
}else{
tmp=tmp->_next;
prev=prev->_next;
}
}
return head->_next;
}
//构造一个链表
node_p create_list(int *array,int size)
{
assert(array);
node dummy(-1);
node_p prev=&dummy;
node_p head=prev;
for(int i=0;i<size;++i){
node_p Node=new node (array[i]);
prev->_next=Node;
prev=prev->_next;
}
return head->_next;
}
};partition_list.cpp
#include "partition_list.h"
using namespace std;
int main()
{
int array[]={1,4,3,2,5,2};
Solution s1;
node_p list=s1.create_list(array,sizeof(array)/sizeof(array[0]));
node_p newlist=s1.partition(list,4);
node_p tmp=newlist;
while(tmp){
cout<<tmp->_val<<" ";
free(tmp);
tmp=tmp->_next;
}
cout<<endl;
return 0;
}运行结果：
下面为leetcode实现的代码：
《完》
本文出自 “零蛋蛋” 博客，请务必保留此出处http://lingdandan.blog.51cto.com/10697032/1783099
原文地址：http://lingdandan.blog.51cto.com/10697032/1783099
