【16-20】LeetCode：Python解题

16. 3Sum Closest【Medium】

Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

For example, given array S = {-1 2 1 -4}, and target = 1.The sum that is closest to the target is 2. (-1 + 2 + 1 = 2).

Solution:

class Solution(object):    def threeSumClosest(self, nums, target):        """        :type nums: List[int]        :type target: int        :rtype: int        """        nums.sort()        lens = len(nums)        result = 100000000        for i in range(lens-2):            if i > 0 and nums[i-1] == nums[i]:                continue            l, r = i+1, lens-1            while l < r:                s = nums[i] + nums[l] + nums[r]                if s == target:                    return target                if abs(s-target) < abs(result-target):                    result = s                if s > target:                    r -= 1                    while l < r and nums[r] == nums[r+1]:                        r -= 1                 elif s < target:                    l += 1                    while l < r and nums[l] == nums[l-1]:                        l += 1         return result

17. Letter Combinations of a Phone Number【Medium】

Given a digit string, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below.

Input:Digit string "23"Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

Solution:

class Solution(object):    def letterCombinations(self, digits):        """        :type digits: str        :rtype: List[str]        """        dic = {'2':'abc', '3':'def', '4':'ghi', '5':'jkl', '6':'mno', '7':'pqrs', '8':'tuv', '9':'wxyz'}        return [] if not digits else reduce(lambda x, y : [i + j for i in x for j in dic[y]], digits, [''])

18. 4Sum【Medium】

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

Note: The solution set must not contain duplicate quadruplets.

For example, given array S = [1, 0, -1, 0, -2, 2], and target = 0.A solution set is:[  [-1,  0, 0, 1],  [-2, -1, 1, 2],  [-2,  0, 0, 2]]

Solution:

class Solution(object):    def fourSum(self, nums, target):        """        :type nums: List[int]        :type target: int        :rtype: List[List[int]]        """        nums.sort()        lenn = len(nums)        result = []        for i in range(lenn-3):            if i > 0 and nums[i] == nums[i-1]:                continue            for j in range(i+1, lenn):                if j > i+1 and nums[j] == nums[j-1]:                    continue                l, r = j+1, lenn-1                while l < r:                    s = nums[i] + nums[j] + nums[l] +nums[r]                    if s < target:                        l += 1                    elif s > target:                        r -= 1                    else:                        result.append([nums[i], nums[j], nums[l], nums[r]])                         l, r = l+1, r-1                        while l < r and nums[l-1] == nums[l]:                            l += 1                        while l < r and nums[r+1] == nums[r]:                            r -= 1        return result

Discussion:

参考：https://discuss.leetcode.com/topic/22705/python-140ms-beats-100-and-works-for-n-sum-n-2/2

19. Remove Nth Node From End of List【Medium】

Given a linked list, remove the nth node from the end of list and return its head.

For example,

Given linked list: 1->2->3->4->5, and n = 2.   After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will always be valid.
Try to do this in one pass.

Solution:

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def recursion(self, parent, node, n):        if node == None:            return 1        num = self.recursion(node, node.next, n)        if num == n:            if parent == node:                nextNone = node.next                node.val = nextNone.val                node.next = nextNone.next                nextNone = None            else:                parent.next = node.next                node.next = None        return num + 1    def removeNthFromEnd(self, head, n):        """        :type head: ListNode        :type n: int        :rtype: ListNode        """        if head.next == None:            return []        self.recursion(head, head, n)        return head

20. Valid Parentheses【Easy】

Given a string containing just the characters ‘(‘, ‘)’, ‘{‘, ‘}’, ‘[’ and ‘]’, determine if the input string is valid.

The brackets must close in the correct order, “()” and “()[]{}” are all valid but “(]” and “([)]” are not.

Solution:

class Solution(object):    def isValid(self, s):        """        :type s: str        :rtype: bool        """        dic = {')': '(', ']': '[', '}': '{'}        l = []        for i in s:            if not l or i not in dic or (i in dic and l[-1] != dic[i]):                l.append(i)            else:                l.pop(-1)        return False if l else True