[LeetCode] 86. 分隔链表

题目链接 : https://leetcode-cn.com/problems/partition-list/

题目描述:

给定一个链表和一个特定值 x，对链表进行分隔，使得所有小于 x 的节点都在大于或等于 x 的节点之前。

你应当保留两个分区中每个节点的初始相对位置。

示例:

输入: head = 1->4->3->2->5->2, x = 3
输出: 1->2->2->4->3->5

思路:

用两个链表,一个链表放小于x的节点,一个链表放大于等于x的节点

最后,拼接这两个链表.

代码:

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def partition(self, head: ListNode, x: int) -> ListNode:
        dummy1 = ListNode(-1)
        dummy2 = ListNode(-1)
        p1 = dummy1
        p2 = dummy2
        while head:
            if head.val < x:
                p1.next = head
                p1 = p1.next
            else:
                p2.next = head
                p2 = p2.next
            head = head.next
        # print(listNodeToString(dummy1.next))
        # print(listNodeToString(dummy2.next))
        p1.next = dummy2.next
        p2.next = None
        return dummy1.next

java

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode partition(ListNode head, int x) {
        ListNode dummy1 = new ListNode(-1);
        ListNode dummy2 = new ListNode(-1);
        ListNode p1 = dummy1;
        ListNode p2 = dummy2;
        while (head != null) {
            if (head.val < x) {
                p1.next = head;
                p1 = p1.next;
            } else {
                p2.next = head;
                p2 = p2.next;
            }
            head = head.next;
        }
        p1.next = dummy2.next;
        p2.next = null;
        return dummy1.next;
    }
}

原文地址：https://www.cnblogs.com/powercai/p/10994134.html