leetcode -- Add Binary

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

For example,
a = "11"
b = "1"
Return "100".

[解题思路]

将a,b转成10进制相加得到结果再转成二进制，可以过small，跑large时溢出

"10100000100100110110010000010101111011011001101110111111111101000000101111001110001111100001101", "110101001011101110001111100110001010100001101011101010000011011011001011101111001100000011011110011" "11101000101011001000011011000001100011110011010010011000000000" "110111101100010011000101110110100000011101000101011001000011011000001100011110011010010011000000000" 
 1 public class Solution {
 2     public String addBinary(String a, String b) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         if(a == null || a.equals("")){
 6             return b;
 7         }
 8         if(b == null || b.equals("")){
 9             return a;
10         }
11         
12         long a1 = 0;
13         for(int i = 0; i < a.length(); i++){
14             a1 = a1 * 2 + (a.charAt(i) - '0');
15         }
16         long b2 = 0;
17         for(int i = 0; i < b.length(); i++){
18             b2 = b2 * 2 + (b.charAt(i) - '0');
19         }
20         
21         long sum = a1 + b2;
22         if(sum == 0){
23             return "0";
24         }
25         ArrayList<Long> binarySum = new ArrayList<Long>();
26         while(sum > 0){
27             binarySum.add(sum % 2);
28             sum = sum / 2;
29         }
30         StringBuilder sb = new StringBuilder();
31         for(int i = binarySum.size() - 1; i >= 0; i-- ){
32             sb.append(binarySum.get(i));
33         }
34         return sb.toString();
35     }
36 }

字符串操作

1. StringBuilder.reverse()

2. 最后一位的进位需要加上

 1 public class Solution {
 2     public String addBinary(String a, String b) {
 3         // Start typing your Java solution below
 4         // DO NOT write main() function
 5         if(a == null || a.equals("")){
 6             return b;
 7         }
 8         if(b == null || b.equals("")){
 9             return a;
10         }
11         StringBuilder a1 = new StringBuilder(a);
12         a1.reverse();
13         StringBuilder b1 = new StringBuilder(b);
14         b1.reverse();
15         StringBuilder result = new StringBuilder();
16         int len = Math.max(a.length(), b.length());
17         int carry = 0;
18         for(int i = 0; i < len; i++){
19             int t1 = (i >= a.length() ? 0 : (a1.charAt(i) - '0'));
20             int t2 = (i >= b.length() ? 0 : (b1.charAt(i) - '0'));
21             int t3 = t1 + t2 + carry;
22             carry = t3 / 2;
23             t3 = t3 % 2;
24             result.append(t3);
25         }
26         if(carry != 0)
27             result.append(carry);
28         result.reverse();
29         return result.toString();
30     }
31 }

 

posted @ 2013-08-13 10:13  feiling  阅读(371)  评论(0)  编辑  收藏  举报