Leetcode Hacking Practice -- Solution in Python Part 1 (發表於10年前)

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All solutions are in Python！

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S算法問題: Add Binary
問題描述:
Given two binary strings, return their sum (also a binary string).
For example,
a = "11"
b = "1"
Return "100".

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算法問題推薦解法來了...

'1'-'0' = 1.

解法（Python）

代碼:


class Solution:

    def addBinary(self, a, b):

        return bin(int(a, 2) + int(b, 2)).split('b')[1]



    def addBinary(self, a, b):

       """Sometimes built-in function cheats too much.

       """

        res, carry, len_a, len_b, i = "", 0, len(a), len(b), 0

        for i in range(max(len_a, len_b)):

            sum = carry

            if i < len_a:

                sum += int(a[-(i + 1)])

            if i < len_b:

                sum += int(b[-(i + 1)])

            carry = sum / 2

            sum = sum % 2

            res = "{0}{1}".format(sum, res)

        if carry == 1:

            res = "1" + res

        return res



    # def addBinary(self, a, b):

    #    """Using carry without sum is also fine. But the readability sucks.

    #    """

    #     res, carry, len_a, len_b, i = "", 0, len(a), len(b), 0

    #     for i in range(max(len_a, len_b)):

    #         if i < len_a:

    #             carry += int(a[-(i + 1)])

    #         if i < len_b:

    #             carry += int(b[-(i + 1)])

    #         res = "{0}{1}".format(carry % 2, res)

    #         carry = carry / 2

    #     if carry == 1:

    #         res = "1" + res

    #     return res

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S算法問題: Add Two Numbers
問題描述:
You are given two linked lists representing two non-negative numbers.
The digits are stored in reverse order and each of their nodes contain a single digit.
Add the two numbers and return it as a linked list.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

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算法問題推薦解法來了...

dummy head...

解法（Python）

代碼:


class Solution:

    def addTwoNumbers(self, l1, l2):

        dummy = ListNode(0)

        current, carry = dummy, 0

        while l1 != None or l2 != None:

            res = carry

            if l1 != None:

                res += l1.val

                l1 = l1.next

            if l2 != None:

                res += l2.val

                l2 = l2.next

            carry, res = res / 10, res % 10

            current.next = ListNode(res)

            current = current.next

        if carry == 1:

            current.next = ListNode(1)

        return dummy.next

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S算法問題: Anagrams
問題描述:
Given an array of strings, return all groups of strings that are anagrams.
Note: All inputs will be in lower-case.

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算法問題推薦解法來了...

Sort the string to see if they're anagrams.
Solution 1 is simpler than 2.

解法（Python）

代碼:


class Solution:

    def anagrams(self, strs):

        anagram_map, res = {}, []

        for str in strs:

            sorted_str = ("").join(sorted(str))

            if sorted_str in anagram_map:

                anagram_map[sorted_str].append(str)

            else:

                anagram_map[sorted_str] = [str]

        for anagrams in anagram_map.values():

            if len(anagrams) > 1:

                res += anagrams

        return res



    # def anagrams(self, strs):

    #     """List comprehension may be more elegant but less readable here.

    #     """

    #     anagram_map = {}

    #     for str in strs:

    #         sorted_str = ("").join(sorted(str))

    #         if sorted_str in anagram_map:

    #             anagram_map[sorted_str].append(str)

    #         else:

    #             anagram_map[sorted_str] = [str]

    #     return [anagram for anagrams in anagram_map.values() if len(anagrams) > 1 for anagram in anagrams]

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S算法問題: Balanced Binary Tree
問題描述:
Given a binary tree, determine if it is height-balanced.
For this problem, a height-balanced binary tree is defined as a binary tree in which
the depth of the two subtrees of every node never differ by more than 1.

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算法問題推薦解法來了...

DFS.

解法（Python）

代碼:


class Solution:

    def isBalanced(self, root):

        return self.getHeight(root) != -1

    

    def getHeight(self, root):

        if root == None:

            return 0

        left_height, right_height = self.getHeight(root.left), self.getHeight(root.right) 

        if left_height < 0 or right_height < 0 or math.fabs(left_height - right_height) > 1:

            return -1

        return max(left_height, right_height) + 1

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S算法問題: Best Time to Buy and Sell Stock II
問題描述:
Say you have an array for which the ith element is the price of a given stock on day i.
Design an algorithm to find the maximum profit. You may complete as many transactions as you like
(ie, buy one and sell one share of the stock multiple times).
However, you may not engage in multiple transactions at the same time
(ie, you must sell the stock before you buy again).