[LeetCode]Partition List
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题目:Partition List
给定一个链表和数值x,保证小于该数值x的都在他左边,其他的相对位置不能改变。
思路:
遍历一边链表,小于x的节点连在一起,其他的连在一起,不改变他们的顺序在首尾相连就可以了。
package com.example.medium; /** * 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. * @author FuPing * */ public class Partition { public class ListNode { int val; ListNode next; ListNode(int x) { val = x; } } public ListNode partition(ListNode head, int x) { ListNode lessList = null,moreList = null; ListNode lessTail = null,moreTail = null; while(head != null){ if(head.val < x){ if(lessList == null){ lessList = head; lessTail = head; }else{ lessTail.next = head; lessTail = lessTail.next; } }else{ if(moreList == null){ moreList = head; moreTail = head; }else{ moreTail.next = head; moreTail = moreTail.next; } } head = head.next; } if(lessList == null)return moreList; lessTail.next = moreList; if(moreList != null)moreTail.next = null; return lessList; } public static void main(String[] args) { // TODO Auto-generated method stub } }
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