Archive for November, 2017

Reverse Integer

Given a 32-bit signed integer, reverse digits of an integer.

Example 1:

Input: 123
Output: 321
Example 2:

Input: -123
Output: -321
Example 3:

Input: 120
Output: 21
Assume we are dealing with an environment which could only hold integers within the 32-bit signed integer range. For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.

class Solution:
    def reverse(self, x):
        :type x: int
        :rtype: int
        list_x = []
        if x >0 :
            for str_x in str(x).split():
            r_list_x = list_x[::-1]
            for item in list_x:
                r_str_x =  "".join(item)
                return (int (r_str_x))    

What is wrong with this one:
x = 123,

str(x).split() is still “123”

split is split the spaces…


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two sum V3

class Solution:
    def twoSum(self, nums, target):
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        dic_value_index = {}
        index_1 = int ()
        index_2 = int ()
        for index, value in enumerate(nums):
            dic_value_index [value] = index
        for item in nums:
            sec_num = target - item
            if sec_num in dic_value_index:
                if nums.index(item) != dic_value_index[sec_num]: # if they are not the same number
                    index_1 = nums.index(item)
                    index_2 = dic_value_index[sec_num]
        return ([index_1, index_2])

Key of this solution:
1. enumerate (list_name) will walk through the index and the value;
2. put the value and index into the dictinary
3. search whether the sec_num in the dictionary (only key can be searched? )
4. if yes, the the value is the index of the sec_num
5. check whether the sec_num is actually the first number, if not , then return the index.

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Two Sum V2

class Solution:
    def twoSum(self, nums, target):
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        nums_length = len (nums) 
        index_1 = int ()
        index_2 = int ()
        target_list = []
        for x in range (nums_length):
            second_num = target - nums[x]
            if second_num != nums[x]:
                if second_num in nums:
                    index_2 = nums.index(second_num)
                    index_1 = x
            elif second_num in nums[x+1::]:
                index_2 = nums[x+1::].index(second_num)+x +1
                index_1 = x
        target_list = [index_1, index_2]
        return target_list

The concept in this solution:
1. have the first number,
2. second_number = target – first_number
3. check whether the second_number in list.
4. if not, then move to next number in the list
5. if yes, check whether the first number and the second number is the same: if it is the same, then make sure it is two different numbers , but starting a new list nums[x+1 ::], and then find the index in the new list. The important thing here is: for the new list starting index is not “0”, it should be “x+1”!!
6. if the second number is not the first number ,then it is easy to just identify the index.

Lesson learned:

1. how to use the value to identify the index: list_name.index(the_value)
2. the second list starting index for nums[x+1::] is x+1 , for example nums[x+1::] [y], then the original index is (x+1) +y

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Leetcode Two Sum – solution – V1

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.


Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

My first version:


nested two for loops, find the two value sum == target


My second version:

also two loops: the first loop is the same, but the second loop starts from x+1, no need to start index 0.

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