[Leetcode 28] 67 Add Binary
Problem:
Given two binary strings, return their sum (also a binary string).
For example,
a = "11"
b = "1"
Return "100"
.
Analysis:
Three main steps: 1. add a and b until meet one's end; 2. process the remaining part of the the other string; 3. process the carry bit.
Simulation problem, the time complexity is O(n+m), the space complexity is O(n+m)
Code:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 class Solution { 2 public: 3 string addBinary(string a, string b) { 4 // Start typing your C/C++ solution below 5 // DO NOT write int main() function 6 int la = a.length()-1, lb = b.length()-1; 7 int bitSum, carry = 0, idx=0; 8 char *res = new char[100](); 9 10 while (la>=0 && lb>=0) { 11 bitSum = a[la] + b[lb] + carry - '0' - '0'; 12 if (bitSum == 2) {//there is a carry 13 res[idx++] = '0'; 14 carry = 1; 15 } else if (bitSum == 3) { 16 res[idx++] = '1'; 17 carry = 1; 18 } else { 19 res[idx++] = bitSum + '0'; 20 carry = 0; 21 } 22 23 la--; 24 lb--; 25 } 26 27 while (la>=0) { 28 bitSum = a[la] + carry - '0'; 29 if (bitSum == 2) {//there is a carry 30 res[idx++] = '0'; 31 carry = 1; 32 } else { 33 res[idx++] = bitSum + '0'; 34 carry = 0; 35 } 36 37 la--; 38 } 39 40 while (lb>=0) { 41 bitSum = b[lb] + carry - '0'; 42 if (bitSum == 2) {//there is a carry 43 res[idx++] = '0'; 44 carry = 1; 45 } else { 46 res[idx++] = bitSum + '0'; 47 carry = 0; 48 } 49 50 lb--; 51 } 52 53 if (carry > 0) res[idx++] = carry + '0'; 54 res[idx] = '\0'; 55 56 for (int i=0, j=idx-1; i <=j; i++, j--) { 57 char ch = res[i]; 58 res[i] = res[j]; 59 res[j] = ch; 60 } 61 62 string sres = res; 63 return sres; 64 } 65 };
Attention: