Jan 23 - Partition List; Linked List; Pointer;-优快云博客

本文详细解析了链表分区算法的实现过程，并分享了一个初始错误的理解及修正后的正确实现方式。通过对链表中小于给定值x的节点进行重新排列，保持原有相对顺序不变的同时，实现了有效的分区操作。

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At the first sight of the problem, I misunderstood it as sort the former part of the list and then keep the original order of those nodes in the larger or equal to target value part. Thus, I sort the first part and get a wrong result.

Code:

/**
 * 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) {
        ListNode cur = head;
        ListNode head1 = null, tail1 = null;
        ListNode head2 = null, tail2 = null;
        while(cur != null){
            ListNode nex = cur.next;
            ListNode node = cur;
            if(node.val < x){
                ListNode cur1 = head1; 
                ListNode prev1 = null;
                while(cur1 != null && cur1.val < node.val) {
                    prev1 = cur1;
                    cur1 = cur1.next;
                }
                if(prev1 != null) prev1.next = node;
                else head1 = node;
                cur.next = cur1;
                if(cur1 == null) tail1 = node;
            }
            else{
                ListNode cur2 = tail2;
                if(cur2 != null){
                    tail2.next = node;
                    node.next = null;
                    tail2 = node;
                }
                else{
                    head2 = node;
                    tail2 = node;
                    node.next = null;
                }
            }
            cur = nex;
        }
        if(tail1 != null) tail1.next = head2;
        if(head1 == null) return head2;
        return head1;
    }
}

After a littile modification, we can the the right one:

/**
 * 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) {
        ListNode cur = head;
        ListNode head1 = null, tail1 = null;
        ListNode head2 = null, tail2 = null;
        while(cur != null){
            ListNode nex = cur.next;
            ListNode node = cur;
            if(node.val < x){
                ListNode cur1 = tail1;
                if(cur1 != null){
                    tail1.next = node;
                    node.next = null;
                    tail1 = node;
                }
                else{
                    head1 = node;
                    tail1 = node;
                    node.next = null;
                }
            }
            else{
                ListNode cur2 = tail2;
                if(cur2 != null){
                    tail2.next = node;
                    node.next = null;
                    tail2 = node;
                }
                else{
                    head2 = node;
                    tail2 = node;
                    node.next = null;
                }
            }
            cur = nex;
        }
        if(tail1 != null) tail1.next = head2;
        if(head1 == null) return head2;
        return head1;
    }
}