コード例 #1
0
        dummy = ListNode(-1)
        dummy.next = head

        prev_node = dummy

        while head and head.next:
            # Nodes to be swapped
            first_node = head
            second_node = head.next

            # Swapping
            prev_node.next = second_node
            first_node.next = second_node.next
            second_node.next = first_node

            # Reinitializing the head and prev_node for next swap
            prev_node = first_node
            head = first_node.next

        # Return the new head node.
        return dummy.next


if __name__ == '__main__':
    head = generateList([1, 2, 3, 4, 5])
    solution = Solution()
    printList(solution.swapPairs_1(head))
    head = generateList([1, 2, 3, 4, 5])
    printList(solution.swapPairs_2(head))

コード例 #2
0
# Definition for singly-linked list.
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None


class Solution:
    def removeElements(self, head: ListNode, val: int) -> ListNode:
        sentinel = ListNode(0)
        sentinel.next = head

        prev, curr = sentinel, head
        while curr:
            if curr.val == val:
                prev.next = curr.next
            else:
                prev = curr
            curr = curr.next

        return sentinel.next


if __name__ == '__main__':
    head = generateList([1, 2, 6, 3, 4, 5, 6])
    val = 6

    solution = Solution()
    printList(solution.removeElements(head, val))
コード例 #3
0
        # Restore the list and return the result.
        first_half_end.next = self.reverse_list(second_half_start)
        return result

    def end_of_first_half(self, head):
        fast = head
        slow = head
        while fast.next is not None and fast.next.next is not None:
            fast = fast.next.next
            slow = slow.next
        return slow

    def reverse_list(self, head):
        previous = None
        current = head
        while current is not None:
            next_node = current.next
            current.next = previous
            previous = current
            current = next_node
        return previous


if __name__ == '__main__':
    head = generateList([1, 2, 3, 3, 2, 1])

    solution = Solution()
    print(solution.isPalindrome_1(head))
    print(solution.isPalindrome_2(head))
    print(solution.isPalindrome_3(head))
コード例 #4
0
ファイル: mergeTwoLists.py プロジェクト: MaoningGuan/LeetCode
    def __init__(self, x):
        self.val = x
        self.next = None


class Solution:
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        pre = ListNode(0)
        head = pre
        while l1 and l2:
            if l1.val <= l2.val:
                pre.next = l1
                l1 = l1.next
            else:
                pre.next = l2
                l2 = l2.next
            pre = pre.next
        pre.next = l1 if l1 else l2
        return head.next


if __name__ == '__main__':
    l1 = generateList([1, 2, 4])
    l2 = generateList([1, 3, 4])

    solution = Solution()
    printList(l1)
    printList(l2)
    mergeList = solution.mergeTwoLists(l1, l2)
    printList(mergeList)
コード例 #5
0
ファイル: deleteNode.py プロジェクト: MaoningGuan/LeetCode
        """
        单指针遍历
        间复杂度 O(N) : N 为链表长度,删除操作平均需循环 N/2 次,最差 N 次。
        空间复杂度 O(1) : pre 占用常数大小额外空间。
        :param head:
        :param val:
        :return:
        """
        if not head:
            return

        if head.val == val:
            return head.next

        pre = ListNode(0)
        pre.next = head
        while pre.next:
            if pre.next.val == val:
                pre.next = pre.next.next
                break
            pre = pre.next
        return head


if __name__ == '__main__':
    head = generateList([-3, 5, -99])
    val = -3

    solution = Solution()
    printList(solution.deleteNode(head, val))
コード例 #6
0
ファイル: app.py プロジェクト: lukegerhart/mediaserver
def home():
    dir = generateList(basedir)
    return render_template('index.html', dir=dir)
コード例 #7
0
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next


class Solution:
    def oddEvenList(self, head: ListNode) -> ListNode:
        if not head:
            return None

        odd = head
        even = head.next
        evenHead = even

        while even and even.next:
            odd.next = even.next
            odd = odd.next
            even.next = odd.next
            even = even.next

        odd.next = evenHead

        return head


if __name__ == '__main__':
    head = generateList([1, 2, 3, 4, 5, 6, 7, 8, 9])
    solution = Solution()

    printList(solution.oddEvenList(head))
コード例 #8
0
        :param head:
        :return:
        """
        if not head or not head.next: return head  # termination.
        # cut the LinkedList at the mid index.
        slow, fast = head, head.next
        while fast and fast.next:
            fast, slow = fast.next.next, slow.next
        mid, slow.next = slow.next, None  # save and cut.
        # recursive for cutting.
        left, right = self.sortList(head), self.sortList(mid)
        # merge `left` and `right` linked list and return it.
        h = res = ListNode(0)
        while left and right:
            if left.val < right.val:
                h.next, left = left, left.next
            else:
                h.next, right = right, right.next
            h = h.next
        h.next = left if left else right
        return res.next


if __name__ == '__main__':
    nums = [4, 2, 1, 3]
    head = generateList(nums)
    printList(head)

    solution = Solution()
    printList(solution.sortList(head))
コード例 #9
0
ファイル: reversePrint.py プロジェクト: MaoningGuan/LeetCode
# Definition for singly-linked list.
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None


class Solution:
    def reversePrint_1(self, head: ListNode) -> List[int]:
        return self.reversePrint_1(head.next) + [head.val] if head else []

    def reversePrint_2(self, head: ListNode) -> List[int]:
        stack = list()
        while head:
            stack.append(head.val)
            head = head.next
        stack.reverse()
        # return stack
        return stack[::-1]


if __name__ == '__main__':
    head = generateList([1, 3, 2, 5, 9, 0])
    solution = Solution()
    print(solution.reversePrint_1(head))
    printList(head)
    print(solution.reversePrint_2(head))
    printList(head)

コード例 #10
0
        :return:
        """
        prehead = ListNode(-1)

        prev = prehead
        while l1 and l2:
            if l1.val <= l2.val:
                prev.next = l1
                l1 = l1.next
            else:
                prev.next = l2
                l2 = l2.next
            prev = prev.next

        # 合并后 l1 和 l2 最多只有一个还未被合并完,我们直接将链表末尾指向未合并完的链表即可
        prev.next = l1 if l1 is not None else l2

        return prehead.next


if __name__ == '__main__':
    list_node_1 = generateList([1, 2, 4])
    list_node_2 = generateList([1, 3, 4])

    solution = Solution()

    # printList(solution.mergeTwoLists_1(list_node_1, list_node_2))
    printList(solution.mergeTwoLists_2(list_node_1, list_node_2))


コード例 #11
0
        l1 = self.merge(lists, left, mid)
        l2 = self.merge(lists, mid + 1, right)
        return self.mergeTwoLists(l1, l2)

    def mergeTwoLists(self, l1, l2):
        if not l1: return l2
        if not l2: return l1
        if l1.val < l2.val:
            l1.next = self.mergeTwoLists(l1.next, l2)
            return l1
        else:
            l2.next = self.mergeTwoLists(l1, l2.next)
            return l2


if __name__ == '__main__':
    soltuion = Solution()

    listNode1 = generateList([1, 4, 5])
    listNode2 = generateList([1, 3, 4])
    listNode3 = generateList([2, 6])

    list_nodes = [listNode1, listNode2, listNode3]

    mergeList1 = soltuion.mergeKLists(list_nodes)
    printList(mergeList1)

    mergeList2 = soltuion.mergeKLists_2(list_nodes)
    printList(mergeList2)