Partition List - LeetCode

最新推荐文章于 2025-01-26 09:55:47 发布

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本文详细介绍了如何通过链表和给定值x将链表分为两部分，一部分包含所有小于x的节点，另一部分包含所有大于等于x的节点，同时保持原始相对顺序不变。

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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.

/**
 * 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(ListNode head, int x) {
	        if(head == null){
	        return null;
	        }
	        ListNode big = new ListNode(0);
	        ListNode bigcurrent = new ListNode(0);
	        big = bigcurrent;
	        ListNode small = new ListNode(0);
	        ListNode smallhead = new ListNode(0);
	        smallhead = small;
	        ListNode current = new ListNode(0);
	        current = head;
	        while(current != null){
	            if(current.val < x){
	                small.next = new ListNode(current.val);
	                small = small.next;
	            }
	            else{
	                bigcurrent.next =  new ListNode(current.val);
	                bigcurrent = bigcurrent.next;
	                }
	            current = current.next;
	        }
	        small.next = big.next;
	        return smallhead.next;
    }
}