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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.
Analysis:
1. build two empty lists
2. val < x add it to the tail of first list
val >= x add it to the tail of second list
3. concatenate the tail of list1 to the head of list2
4. the tail of list2 list.next = null
Java code:
20160601
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode partition(ListNode head, int x) { //1. build two empty lists //2. val < x add it to the tail of first list // val >= x add it to the tail of second list //3. concatenate the tail of list1 to the head of list2 //4. the tail of list2 list.next = null //base case if(head == null || head.next == null) { return head; } ListNode cur = head; ListNode dummy1 = new ListNode(0); ListNode head1 = dummy1; ListNode dummy2 = new ListNode(0); ListNode head2 = dummy2; while( cur != null) { if (cur.val < x) { dummy1.next = cur; dummy1 = dummy1.next; } else { dummy2.next = cur; dummy2 = dummy2.next; } cur = cur.next; } dummy2.next = null; dummy1.next = head2.next; return head1.next; } }
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原文地址：http://www.cnblogs.com/anne-vista/p/5551725.html
