LeetCode 答案（python）18-24

18.四个数之和

给定一个包含 n 个整数的数组 nums 和一个目标值 target，判断 nums 中是否存在四个元素 a，b，c 和 d ，使得 a + b + c + d 的值与 target 相等？找出所有满足条件且不重复的四元组。

注意：

答案中不可以包含重复的四元组。

示例：

给定数组 nums = [1, 0, -1, 0, -2, 2]，和 target = 0。

满足要求的四元组集合为：
[
  [-1,  0, 0, 1],
  [-2, -1, 1, 2],
  [-2,  0, 0, 2]
]

class Solution:
    def fourSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[List[int]]
        """
        result = []
        N = len(nums)
        if N < 4:
            return result
        nums = sorted(nums)
        for i in range(N - 3):
            if sum(nums[i:i + 4]) > target or sum(nums[-4:]) < target:
                break
            if nums[i] + sum(nums[-3:]) < target:
                continue
            if i > 0 and nums[i] == nums[i - 1]:
                continue
            target2 = target - nums[i]
            for j in range(i + 1, N - 2):
                if sum(nums[j:j + 3]) > target2 or sum(nums[-3:]) < target2:
                    break
                if nums[j] + sum(nums[-2:]) < target2:
                    continue
                if j > i + 1 and nums[j] == nums[j - 1]:
                    continue
                target3 = target2 - nums[j]
                left = j + 1
                right = N - 1
                while (left < right):
                    if nums[left] + nums[right] == target3:
                        result.append([nums[i], nums[j], nums[left], nums[right]])
                        while left < right and nums[left] == nums[left + 1]: left += 1;
                        while left < right and nums[right] == nums[right - 1]: right -= 1;
                        left += 1
                        right -= 1
                    elif nums[left] + nums[right] < target3:
                        left += 1
                    else:
                        right -= 1
        return result

19 删除链表中倒数第n个节点

class Solution:
    def removeNthFromEnd(self, head, n):
        """
        :type head: ListNode
        :type n: int
        :rtype: Li
        """
        dao_n, first = head
        for i in range(n):
            first = first.next
        if not first:
            return dao_n.next
        while first:
            dao_n = dao_n.next
            first = first.next
        dao_n.next = dao_n.next.next
        return head

　　

20 有效括号

给定一个只包括 '('，')'，'{'，'}'，'['，']' 的字符串，判断字符串是否有效。

有效字符串需满足：

1. 左括号必须用相同类型的右括号闭合。
2. 左括号必须以正确的顺序闭合。

注意空字符串可被认为是有效字符串。

示例 1:

输入: "()"
输出: true

示例 2:

输入: "()[]{}"
输出: true

class Solution:
    def isValid(self, s):
        """
        :type s: str
        :rtype: bool
        """
        stack = []
        dict = {"]": "[", "}": "{", ")": "("}
        for char in s:
            if char in dict.values():
                stack.append(char)
            elif char in dict.keys():
                if stack == [] or dict[char] != stack.pop():
                    return False
            else:
                return False
        return stack == []

21合并两个有序链表

class Solution:
    def mergeTwoLists(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        dumy = ListNode(0)
        cur = dumy
        while l1 or l2:
            if l1 and l2:
                if l1.val > l2.val:
                    cur.next = l2
                    l2 = l2.next
                else:
                    cur.next = l1
                    l1 = l1.next
            elif l1:
                cur.next = l1
                l1 = l1.next
            else:
                cur.next = l2
                l2 = l2.next
            cur = cur.next
        return dumy.next

　　

22 括号生成　　

class Solution:
    def generateParenthesis(self, n):
        """
        :type n: int
        :rtype: List[str]22
        """
        if n == 0:
            return []
        left = right = n
        result = []
        self.generate(left, right, result, '')
        return result
    def generate(self, left, right, result, string):
        if left == 0 and right == 0:
            result.append(string)
            return
        if left:
            self.generate(left - 1, right , result, string+'(')
        if left < right:
            self.generate(left, right - 1, result, string+')')

a = Solution()
print(a.generateParenthesis(4))

23合并K个排序链表

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None
class Solution:
    def mergeKLists(self, lists):
        pre = cur = ListNode(0)

        heap = []
        for i in range(len(lists)):
            if lists[i]:
                heapq.heappush(heap, (lists[i].val, i, lists[i]))

        while heap:
            node = heapq.heappop(heap)
            idx = node[1]
            cur.next = node[2]
            cur = cur.nextb

            if cur.next:
                heapq.heappush(heap, (cur.next.val, idx, cur.next))

        return pre.next