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  • LeetCode 86. 分隔链表

    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

    # 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:
            if head is None or head.next is None:
                return head 
            left_head = ListNode(None)
            right_head = ListNode(None)
            cur_left = left_head
            cur_right = right_head
            cur = head 
            while cur:
                if cur.val < x:
                    cur_left.next = cur 
                    cur_left = cur_left.next
                else:
                    cur_right.next = cur
                    cur_right = cur_right.next
                cur = cur.next 
            cur_left.next = right_head.next 
            cur_right.next = None 
            return left_head.next
    
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  • 原文地址:https://www.cnblogs.com/sandy-t/p/13287425.html
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