LeetCode题解(0086):依据特定值分隔链表(Python)

题目:原题链接(中等)

标签:链表、链表-双指针

解法 时间复杂度 空间复杂度 执行用时
Ans 1 (Python) O ( N ) O(N) O ( 1 ) O(1) 36ms (95.15%)
Ans 2 (Python)
Ans 3 (Python)

LeetCode的Python执行用时随缘,只要时间复杂度没有明显差异,执行用时一般都在同一个量级,仅作参考意义。

解法一(双指针):

def partition(self, head: ListNode, x: int) -> ListNode:
    headA = nodeA = ListNode(0)
    headB = nodeB = ListNode(0)
    while head:
        if head.val < x:
            nodeA.next = ListNode(head.val)
            nodeA = nodeA.next
        else:
            nodeB.next = ListNode(head.val)
            nodeB = nodeB.next
        head = head.next
    nodeA.next = headB.next
    return headA.next

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转载自blog.csdn.net/Changxing_J/article/details/107317054