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PartitionList.java
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70 lines (67 loc) · 1.95 KB
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/*
Author: King, higuige@gmail.com
Date: Oct 7, 2014
Problem: Partition List
Difficulty: Easy
Source: https://oj.leetcode.com/problems/partition-list/
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.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
public ListNode partition_1(ListNode head, int x) {
ListNode leftdummy = new ListNode(-1);
ListNode rightdummy = new ListNode(-1);
ListNode lhead = leftdummy;
ListNode rhead = rightdummy;
for(ListNode cur = head; cur != null; 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 = null;
return leftdummy.next;
}
public ListNode partition(ListNode head, int x) {
ListNode dummy = new ListNode(-1);
dummy.next = head;
ListNode cur = dummy;
ListNode ins = dummy;
while (cur.next != null && cur.next.val < x) {
cur = cur.next;
ins = ins.next;
}
while (cur.next != null) {
if (cur.next.val >= x) {
cur = cur.next;
} else {
ListNode node = cur.next;
cur.next = cur.next.next;
node.next = ins.next;
ins.next = node;
ins = ins.next;
}
}
return dummy.next;
}
}