コード例 #1
0
    def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
        dummy = ListNode(0)
        dummy.next = head

        pre, end = dummy, dummy

        while end.next:
            # 取出待翻转的部分
            i = 0
            while i < k and end:
                end = end.next
                i += 1
            if not end: break

            # 断开链表
            startNode = pre.next
            nextNode = end.next
            end.next = None

            # 处理翻转
            pre.next = self.reverse(startNode)
            # startNode 转到翻转这部分节点的最后了

            # 连接断开的链表
            startNode.next = nextNode

            # 挪动以进行下一组处理
            pre = startNode
            end = pre
        return dummy.next
コード例 #2
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 def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
     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
コード例 #3
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 def sortList(self, head: ListNode) -> ListNode:
     res_ls = []
     while head:
         res_ls.append(head.val)
         head = head.next
     if len(res_ls) == 0:
         return
     res_ls.sort()
     node = ListNode(res_ls[0])
     new_head = node
     for i in res_ls[1:]:
         node.next = ListNode(i)
         node = node.next
     return new_head
コード例 #4
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    def reverseKGroup(self, head: ListNode, k: int) -> ListNode:
        dummy = ListNode(0)
        p = dummy
        while True:
            count = k
            stack = []
            tmp = head
            while count and tmp:
                stack.append(tmp)
                tmp = tmp.next
                count -= 1
            # 注意,目前tmp所在k+1位置
            # 说明剩下的链表不够k个,跳出循环
            if count:
                p.next = head
                break
            # 翻转操作
            while stack:
                p.next = stack.pop()
                p = p.next
            # 与剩下链表连接起来
            p.next = tmp
            head = tmp

        return dummy.next
コード例 #5
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 def reverseListNode(self, head: ListNode) -> ListNode:
     if head.next is None:
         return head
     last = self.reverseListNode(head.next)
     head.next.next = head
     head.next = None
     return last
コード例 #6
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 def swapPairs(self, head: ListNode) -> ListNode:
     if not head or not head.next:
         return head
     newHead = head.next
     head.next = self.swapPairs(newHead.next)
     newHead.next = head
     return newHead
コード例 #7
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 def reverseList(self, head: ListNode) -> ListNode:
     if head.next is None:
         return head
     last = self.reverseList(head.next)
     # 环形链表这里vscode没法展示
     head.next.next = head
     head.next = None
     return last
コード例 #8
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    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        def linked_list_value(l1: ListNode) -> int:
            l1_value = []
            while l1 is not None:
                l1_value.append(str(l1.val))
                l1 = l1.next
            return int("".join(l1_value)[::-1])

        l1_value = linked_list_value(l1)
        # print(l1_value)
        l2_value = linked_list_value(l2)
        # print(l2_value)
        value_ = str(l1_value + l2_value)[::-1]

        dummy_head = ListNode(0)
        t = dummy_head
        for i in value_:
            curr = ListNode(val=int(i))
            t.next = curr
            t = t.next
        return dummy_head.next
コード例 #9
0
# Definition for singly-linked list.
from base.linked_list.ListNode import ListNode


class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        if head.next is None:
            return head
        last = self.reverseList(head.next)
        # 环形链表这里vscode没法展示
        head.next.next = head
        head.next = None
        return last


if __name__ == '__main__':
    nums = [1, 2, 3, 4, 5]
    ln = ListNode.list2node(nums)
    sol = Solution()
    a = sol.reverseList(ln)
    print(a)
コード例 #10
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from base.linked_list.ListNode import ListNode


class Solution:
    def kthToLast(self, head: ListNode, k: int) -> int:
        # 快慢指针的解法
        advance, last = head, head
        a = 0
        while a < k:
            advance = advance.next
            a += 1
        while advance:
            advance = advance.next
            last = last.next
        return last.val


if __name__ == '__main__':
    ls = ListNode.list2node([1, 2, 3, 4, 5])
    sol = Solution()
    print(sol.kthToLast(ls, 2))
コード例 #11
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from base.linked_list.ListNode import ListNode


class Solution(object):
    def deleteDuplicates(self, head):
        if not head or not head.next:
            return head
        if head.val != head.next.val:
            head.next = self.deleteDuplicates(head.next)
        else:
            move = head.next
            while move and head.val == move.val:
                move = move.next
            return self.deleteDuplicates(move)
        return head


if __name__ == '__main__':
    head1 = [1, 2, 3, 3, 4, 4, 5]
    head1 = ListNode.list2node(head1)
    head2 = [1, 1, 1, 2, 3]
    head2 = ListNode.list2node(head2)
    sol = Solution()
    print(sol.deleteDuplicates(head2))
    print(sol.deleteDuplicates(head1))
コード例 #12
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# Definition for singly-linked list.
from base.linked_list.ListNode import ListNode


class Solution:
    def getIntersectionNode(self, headA: ListNode,
                            headB: ListNode) -> ListNode:
        reverse_headA = self.reverseListNode(headA)
        reverse_headB = self.reverseListNode(headB)
        while reverse_headA.val == reverse_headB.val:
            reverse_headA = reverse_headA.next
            reverse_headB = reverse_headB.next
        return self.reverseListNode(reverse_headA)

    def reverseListNode(self, head: ListNode) -> ListNode:
        if head.next is None:
            return head
        last = self.reverseListNode(head.next)
        head.next.next = head
        head.next = None
        return last


if __name__ == '__main__':
    a = [1, 2, 3, 4, 5, 6]
    ln = ListNode.list2node(a)
コード例 #13
0
from typing import List
from base.linked_list.ListNode import ListNode


class Solution:
    def reversePrint(self, head: ListNode):
        res = []
        while head is not None:
            res.append(head.val)
            head = head.next
        return res[::-1]


if __name__ == '__main__':
    head = ListNode.list2node([1, 3, 2])
    sol = Solution()
    print(sol.reversePrint(head))
コード例 #14
0
# Definition for singly-linked list.
from base.linked_list.ListNode import ListNode


class Solution:
    def sortList(self, head: ListNode) -> ListNode:
        res_ls = []
        while head:
            res_ls.append(head.val)
            head = head.next
        if len(res_ls) == 0:
            return
        res_ls.sort()
        node = ListNode(res_ls[0])
        new_head = node
        for i in res_ls[1:]:
            node.next = ListNode(i)
            node = node.next
        return new_head


if __name__ == '__main__':
    sol = Solution()
    a = ListNode.list2node([4, 2, 1, 3])
    b = sol.sortList(a)
    print(b)
コード例 #15
0
# Definition for singly-linked list.
from base.linked_list.ListNode import ListNode


class Solution:
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        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__':
    a = [1, 2, 4]
    lna = ListNode.list2node(a)
    print(lna)
    b = [1, 3, 4]
    lnb = ListNode.list2node(b)
    print(lnb)
    sol = Solution()
    print(sol.mergeTwoLists(lna, lnb))
コード例 #16
0
# Definition for singly-linked list.
from base.linked_list.ListNode import ListNode


class Solution:
    def deleteDuplicates(self, head: ListNode) -> ListNode:
        cur = head
        while cur:
            while cur.next and cur.next.val == cur.val:
                cur.next = cur.next.next
            cur = cur.next
        return head


if __name__ == '__main__':
    ln = ListNode.list2node([1, 1, 2])
    ln2 = ListNode.list2node([1, 1, 2, 3, 3])
    sol = Solution()
    print(sol.deleteDuplicates(ln))
    print(sol.deleteDuplicates(ln2))
コード例 #17
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class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        def linked_list_value(l1: ListNode) -> int:
            l1_value = []
            while l1 is not None:
                l1_value.append(str(l1.val))
                l1 = l1.next
            return int("".join(l1_value)[::-1])

        l1_value = linked_list_value(l1)
        # print(l1_value)
        l2_value = linked_list_value(l2)
        # print(l2_value)
        value_ = str(l1_value + l2_value)[::-1]

        dummy_head = ListNode(0)
        t = dummy_head
        for i in value_:
            curr = ListNode(val=int(i))
            t.next = curr
            t = t.next
        return dummy_head.next


if __name__ == '__main__':
    l1 = ListNode.list2node([2, 4, 3])
    l2 = ListNode.list2node([5, 6, 4])
    sol = Solution()
    temp = (sol.addTwoNumbers(l1, l2))
    print(temp)
コード例 #18
0
# Definition for singly-linked list.
from base.linked_list.ListNode import ListNode


class Solution:
    def removeDuplicateNodes(self, head: ListNode) -> ListNode:
        if not head:
            return head
        occurred = {head.val}
        pos = head
        # 枚举前驱节点
        while pos.next:
            # 当前待删除节点
            cur = pos.next
            if cur.val not in occurred:
                occurred.add(cur.val)
                pos = pos.next
            else:
                pos.next = pos.next.next
        return head


if __name__ == '__main__':
    sol = Solution()
    head = ListNode.list2node([1, 2, 3, 3, 2, 1])
    head2 = ListNode.list2node([1, 2, 3])
    print(head)
    print(head2)
    print(sol.removeDuplicateNodes(head))
コード例 #19
0
# Definition for singly-linked list.
from base.linked_list.ListNode import ListNode


class Solution:
    def getDecimalValue(self, head: ListNode) -> int:
        res = []
        while head:
            res.append(str(head.val))
            head = head.next
        return int("".join(res), base=2)


if __name__ == '__main__':
    head = ListNode.list2node([1, 0, 1])
    sol = Solution()
    print(sol.getDecimalValue(head))
コード例 #20
0
from base.linked_list.ListNode import ListNode


class Solution:
    def rotateRight(self, head: ListNode, k: int) -> ListNode:
        pass


if __name__ == '__main__':
    sol = Solution()
    head = [1, 2, 3, 4, 5]
    head = ListNode.list2node(head)
    k = 2
    print(sol.rotateRight(head, k))