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PartitionList.h
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PartitionList.h
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/*
Author: Annie Kim, [email protected] : King, [email protected]
Date: Apr 9, 2013
Update: Oct 7, 2014
Problem: Partition List
Difficulty: Easy
Source: http://leetcode.com/onlinejudge#question_86
Notes:
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.
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, int x) {
ListNode leftdummy(-1);
ListNode rightdummy(-1);
ListNode * lhead = &leftdummy;
ListNode * rhead = &rightdummy;
for(ListNode*cur = head; cur; cur=cur->next){
if(cur->val<x){
lhead->next = cur;
lhead = lhead->next;
}else{
rhead->next = cur;
rhead = rhead->next;
}
}
lhead->next = rightdummy.next;
rhead->next = nullptr;
return leftdummy.next;
}
ListNode *partition_1(ListNode *head, int x) {
ListNode dummy(0), *ins = &dummy, *cur = &dummy;
dummy.next = head;
while (cur->next)
{
if (cur->next->val >= x)
{
cur = cur->next;
}
else
{
if (cur == ins)
{
cur = cur->next;
ins = ins->next;
}
else
{
ListNode *move = cur->next;
cur->next = move->next;
move->next = ins->next;
ins->next = move;
ins = move;
}
}
}
return dummy.next;
}
};