第21题 Partition List

最新推荐文章于 2025-02-10 22:44:08 发布

amberFeb

最新推荐文章于 2025-02-10 22:44:08 发布

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本文详细阐述了如何通过链表遍历与比较实现元素值小于指定值x的节点前置，大于等于x的节点后置，同时保持相对顺序不变。

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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.

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(ListNode head, int x) {
        if(head==null || head.next==null) return head;
        ListNode fakeHead= new ListNode(0);
        fakeHead.next = head; 
        ListNode less=fakeHead;
        while(less.next.val<x){
            less=less.next;
            if(less.next==null)
                return fakeHead.next;
        }
        ListNode great=less.next;
        while(great!=null&&great.next!=null){
            if(great.next.val>=x)   great=great.next;
            else{
                ListNode temp = great.next;
                great.next=temp.next;   //remove temp from list
                temp.next=less.next;    //insert temp after less
                less.next=temp;
                less=less.next; 
                //Notice: dont need great=great.next here. 
                //Because the value of great.next has changed, need to check the value of great.next again
            }
        }
        return fakeHead.next;
        
        
    }
}

Note：一般程序有可能对头节点执行增删，插入或改换位置的操作时，需要新建一个fakeHead，指向头结点，来统一对头结点的操作，减少special case，简化代码。