Daily practice (39): Binary summation

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title: Daily Practice (39): Binary Summation

categories:[Swords offer]

tags:[Daily practice]

date: 2022/04/11

---

Daily practice (39): Binary summation

Given two binary strings, return their sum (in binary).

The input is a non-empty string and contains only the numbers 1 and 0.

Example 1:

Input: a = "11", b = "1"

Output: "100"

Example 2:

Input: a = "1010", b = "1011"

Output: "10101"

hint:

Each string consists only of the characters '0' or '1'.

1 <= a.length, b.length <= 10^4

Strings do not contain leading zeros unless they are "0".

Source: LeetCode

Link: https://leetcode-cn.com/problems/add-binary

Method 1: Simulation operation

Thought analysis

Simulate direct calculation, that is, according to a similar decimal calculation method, considering the carry

 string addBinary(string a, string b) {
    //  先倒序
    reverse(a.begin(), a.end());
    reverse(b.begin(), b.end());

    int na = a.size();
    int nb = b.size();
    int n = max(na, nb);

    string res = "";
    int curr = 0;
    for (int i = 0; i < n; ++i)
    {
        curr += i < na ? a[i] == '1' : 0;
        curr += i < nb ? b[i] == '1' : 0;

        res.push_back(curr % 2 ? '1' : '0');
        curr /= 2;
    }
    if (curr)
    {
        res.push_back('1');
    }
    reverse(res.begin(), res.end());
    return res;
}

Method 2: Simulation operation optimization (LeetCode English website boss writing)

Thought analysis

Simulate direct calculation, that is, according to a similar decimal calculation method, considering the carry

 string addBinary(string a, string b) {
    string s;
    s.reserve(a.size() + b.size());
    int c = 0, i = a.size() - 1, j = b.size() - 1;
    while(i >= 0 || j >= 0 || c == 1)
    {
        c += i >= 0 ? a[i--] - '0' : 0;
        c += j >= 0 ? b[j--] - '0' : 0;
        s.push_back((c & 1) + '0');
        c >>= 1;
    }
    reverse(s.begin(), s.end());
    return s;
}