Problem
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"
Solution (Java)
class Solution {
public String addBinary(String a, String b) {
char[] aArray = a.toCharArray();
char[] bArray = b.toCharArray();
StringBuilder sb = new StringBuilder();
int i = aArray.length - 1;
int j = bArray.length - 1;
int carry = 0;
while (i >= 0 || j >= 0) {
int sum = (i >= 0 ? aArray[i] - '0' : 0) + (j >= 0 ? bArray[j] - '0' : 0) + carry;
sb.append(sum % 2);
carry = sum / 2;
if (i >= 0) {
i--;
}
if (j >= 0) {
j--;
}
}
if (carry != 0) {
sb.append(carry);
}
return sb.reverse().toString();
}
}
Solution (Javascript)
/**
* @param {string} a
* @param {string} b
* @return {string}
*/
var addBinary = function(a, b) {
var len1 = a.length;
var len2 = b.length;
var max = Math.max(len1, len2);
var res = '';
var carry = 0;
var val = 0;
for (var i = 0; i < max; i++) {
val = Number(a[len1 - 1 - i] || 0) + Number(b[len2 - 1 - i] || 0) + carry;
carry = Math.floor(val / 2);
res = (val % 2) + res;
}
if (carry) res = 1 + res;
return res;
};
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(1).