1073. Adding Two Negabinary Numbers

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最新推荐文章于 2020-12-15 20:15:38 发布

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本文链接：https://blog.youkuaiyun.com/wilzxu/article/details/90746293

本文介绍了一种解决LeetCode上Adding Two Negabinary Numbers问题的方法，详细解释了如何处理负二进制数的进位问题，给出了具体的示例和代码实现。

1073. Adding Two Negabinary Numbers

方法1:
- 易错点：
- Complexity

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

方法1:

discussion: https://leetcode.com/problems/adding-two-negabinary-numbers/discuss/303795/C%2B%2B-From-Wikipedia

思路：

先讲一个栗子：
arr1 = [1,1,1,1,1]
arr2 = [1,0,1]
每一位对应的是(-2)^4 + (-2) ^3 + (-2)^2 + (-2) ^1 + (-2) ^0。从最后一位开始 1 + 1 = 2, 向前进位，但是由于下一位代表的是-1，所以carry = -1，下一位的和1 + 0 - 1 = 0，carry 恢复0。继续往前1 + 1 = 2，carry = -1，往前1 - 1= 0， carry = 0。最后一位1 + 0 + 0 = 1，carry = 0，结束。

也就是说，每一次向上进位，由于相邻位符号相反，需要进-1而不是1。而当前位也因为后面可能进位-1，0，1，而出现-1，0，1，2。2的情况下进位-1，-1的情况下怎么表示？举例来讲[1, 0, 1] + [1, 0, 1] = [ ,-1, 0]，也就是 2，可以拆成2 = 4 + (-2) = [1, 1, 0]，对比-2 = [1, 0]。所以当sum = - 1，本位 =1 ，进位carry = 1。最后，由于存在-1，有可能本来是1的高位被清零。所以最后还要将高位0清除，比如下面的栗子。

Input:
[1]
[1,1]
Output:
[0,0]
Expected:
[0]

易错点：

小技巧：for loop继续遍历结束或者carry != 0。但是要小心n1 - i 和 n2 - i 变成负数。
pop前缀0。

Complexity

Time complexity: O(max(n1, n2))
Space complexity: O(max(n1, n2))

class Solution {
public:
    vector<int> addNegabinary(vector<int>& arr1, vector<int>& arr2) {
        // if (arr1.size() < arr2.size()) swap(arr1, arr2);
        int n1 = arr1.size(), n2 = arr2.size();
        int carry = 0;
        vector<int> res;
        for (int i = 0; i <= n1 || carry != 0; i++){
            int sum = ((n1 - i) < 0 ? 0 : arr1[n1 - i]) + ((n2 - i) < 0 ? 0: arr2[n2 - i]) + carry;
            res.push_back(abs(sum) % 2);
            if (sum < 0) carry = 1;
            else if (sum > 1) carry = -1;
            else carry = 0;
        }
        while (res.size() > 1 && res.back() == 0) res.pop_back();
        reverse(res.begin(), res.end());
        return res;
    }
};