【leetcode】1073. Adding Two Negabinary Numbers

题目如下：

Given two numbers arr1 and arr2 in base -2, return the result of adding them together.

Each number is given in array format:  as an array of 0s and 1s, from most significant bit to least significant bit.  For example, arr = [1,1,0,1]represents the number (-2)^3 + (-2)^2 + (-2)^0 = -3.  A number arr in array format is also guaranteed to have no leading zeros: either arr == [0] or arr[0] == 1.

Return the result of adding arr1 and arr2 in the same format: as an array of 0s and 1s with no leading zeros.

Example 1:

Input: arr1 = [1,1,1,1,1], arr2 = [1,0,1]
Output: [1,0,0,0,0]
Explanation: arr1 represents 11, arr2 represents 5, the output represents 16.

Note:

1. 1 <= arr1.length <= 1000
2. 1 <= arr2.length <= 1000
3. arr1 and arr2 have no leading zeros
4. arr1[i] is 0 or 1
5. arr2[i] is 0 or 1

解题思路：负二进制的计算问题。首先把arr1和arr2中较短的那个前面补零直到两者长度一样，由于计算会有进位的情况，因为是负进制，最多只能进两位，所以还需要在arr1和arr2前面补两个零。假设用carry表示进位的情况，那么显然carry的取值只能是0，-1，1。0表示不需要进位，-1表示进位的值和当前位数的值的符号不一致，而1表示1致。例如11 + 01，末位的两个1相加需要进位到倒数第二位，由于倒数第二位的符号和末位是不一致的，因此记carry=-1；又01 + 01，末位进位到倒数第三位，两者符号一致，所以carry为1。记v = arr1[i] + arr2[i]，显然v只能是0，1，2三种取值，加上carry的取值，很轻松的可以得出如下关系：

代码如下：

class Solution(object):
    def addNegabinary(self, arr1, arr2):
        """
        :type arr1: List[int]
        :type arr2: List[int]
        :rtype: List[int]
        """
        length = max(len(arr1), len(arr2))
        arr1 = [0]*2 + [0] * (length - len(arr1)) + arr1
        arr2 = [0]*2 + [0] * (length - len(arr2)) + arr2
        res = []
        carry = 0
        for i in range(len(arr1)):
            inx = len(arr1) - i - 1
            v = arr1[inx] + arr2[inx]
            #carry: 0,1,-1 ; v:0,1,2
            if carry == 1 and v == 0:
                res = [1] + res
                carry = 0
            elif carry == 1 and v == 1:
                res = [0] + res
                carry = -1
            elif carry == 1 and v == 2:
                res = [1] + res
                carry = -1
            elif carry == -1 and v == 0:
                res = [1] + res
                carry = 1
            elif carry == -1 and v == 1:
                res = [0] + res
                carry = 0
            elif carry == -1 and v == 2:
                res = [1] + res
                carry = 0
            elif carry == 0 and v <= 1:
                res = [v] + res
            elif carry == 0 and v == 2:
                res = [0] + res
                carry = -1

        while len(res) > 1:
            if res[0] == 0:
                res.pop(0)
                continue
            break

        return res