461. Hamming Distance and 477. Total Hamming Distance in Python

題目：

The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

Given two integers x and y, calculate the Hamming distance.

Note:
0 ≤ x, y < 231.

Example:

Input: x = 1, y = 4
Output: 2
Explanation:
1   (0 0 0 1)
4   (0 1 0 0)
       ↑   ↑
The above arrows point to positions where the corresponding bits are different.

Example:

Input: 4, 14, 2
Output: 6
Explanation: In binary representation, the 4 is 0100, 14 is 1110, and 2 is 0010 (just
showing the four bits relevant in this case). So the answer will be:
HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2 + 2 = 6.

 Note:

1. Elements of the given array are in the range of 0 to 10^9
2. Length of the array will not exceed 10^4

代碼：

這兩週比較慢，都沒有作題，趕忙作兩道。這兩個題目比較像，就放在一塊兒吧。漢明距離，就是把數字變成二進制，看有多少爲不同，不同的個數就是漢明距離。

因此，第一題，僅僅比較兩個數字，我就用了常人都能想到的方法，轉換成二進制字符串，遍歷比較。惟一注意的就是將較短的字符串前面加0補足和長的相同位數。

   def hammingDistance(self, x, y):
        """
        :type x: int
        :type y: int
        :rtype: int
        """
        str1 = bin(x)[2:]
        str2 = bin(y)[2:]
        if len(str1) < len(str2):
            str1 = '0'*(abs(len(str2)-len(str1)))+str1
        else:
            str2 = '0' * (abs(len(str2) - len(str1))) + str2

        res = 0
        for i in range(0,len(str1)):
            if not str1[i:i+1] == str2[i:i+1]:
                res += 1
        return res

這個題經過了，但第二個題目要求給出一個字符串，兩兩求出漢明距離，而後相加。因而，我接着上題的函數，增長了一個遍歷，O(n^2):

 def totalHammingDistance(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums:
            return 0
        res = 0
        new_nums = sorted(nums)
        print sorted(set(nums))
        for i in range(len(new_nums)):
            for j in new_nums[i+1:]:
                print new_nums[i],j
                print self.hammingDistance(new_nums[i],j)
                res += self.hammingDistance(new_nums[i],j)
        return res

邏輯是沒錯，但不用想，leetcode中固然超時，會有一個包含1000個七、8位數字的列表去測試。唉，沒辦法，想不出來，百度了一下。這個漢明距離，其實能夠經過每一個bit位上面數字0的個數和1的個數相乘的結果，相加求得。等於逐一遍歷列表中每一個數字的每一個bit位，全部bit位遍歷一遍。

    def totalHammingDistance2(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        if not nums:
            return 0
        res = 0
        lists = []
        nums.sort()
        for i in nums:
            temp = list(bin(i)[2:])
            temp.reverse()
            lists.append(temp)

        for i in range(len(lists[-1])):
            count_0, count_1 = 0, 0
            for element in lists:
                # print element
                if len(element)-1 < i:
                    count_0 += 1
                elif element[i] == '0':
                    count_0 += 1
                elif element[i] == '1':
                    count_1 += 1
                # print count_0,count_1
            res += count_0 * count_1
        return res

試了下，果真沒問題！：）