[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:

 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 };
View Code

 

 

Attention:

posted on 2013-05-20 07:24  freeneng  阅读(144)  评论(0)  编辑  收藏  举报

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