[LeetCode] 67. Add Binary

Given two binary strings, return their sum (also a binary string).

The input strings are both non-empty and contains only characters 1 or 0.

Example 1:

Input: a = "11", b = "1"
Output: "100"
Example 2:

Input: a = "1010", b = "1011"
Output: "10101"

Constraints:

Each string consists only of '0' or '1' characters.
1 <= a.length, b.length <= 10^4
Each string is either "0" or doesn't contain any leading zero.

二进制求和。影子题415，是十进制的加法。做法甚至代码几乎都是一样的。

题意是给两个用字符串表示的二进制数字，请返回他们的和，也用字符串表示。思路是按位读取数字，从最低位开始两两相加，之后将（加和%2）append到结果集。为什么要将加和%2是因为这是二进制的计算，之后计算carry的时候也是需要除以2而不是除以10。

时间O(n)

空间O(1)

Java实现

class Solution {
    public String addBinary(String a, String b) {
        int i = a.length() - 1;
        int j = b.length() - 1;
        int carry = 0;
        int sum = 0;
        StringBuilder sb = new StringBuilder();
        while (i >= 0 || j >= 0 || carry > 0) {
            int x = i >= 0 ? a.charAt(i--) - '0' : 0;
            int y = j >= 0 ? b.charAt(j--) - '0' : 0;
            sum = x + y + carry;
            sb.append(sum % 2);
            carry = sum / 2;
        }
        return sb.reverse().toString();
    }
}

JavaScript实现

/**
 * @param {string} a
 * @param {string} b
 * @return {string}
 */
var addBinary = function(a, b) {
    let i = a.length - 1;
    let j = b.length - 1;
    let carry = 0;
    let res = '';
    while (i >= 0 || j >= 0 || carry > 0) {
        if (i >= 0) {
            carry += parseInt(a[i]);
        }
        i--;
        if (j >= 0) {
            carry += parseInt(b[j]);
        }
        j--;
        res = (carry % 2) + res;
        carry = parseInt(carry / 2);
    }
    return res;
};

LeetCode 题目总结