标签:style blog http color os io for re
Given two binary strings, return their sum (also a binary string).
For example,
a ="11"
b ="1"
Return"100"
.
算法思路:
模拟二进制加法,跟十进制木有区别,将a,b转置(不转置的话,倒着遍历也是同样道理),这样就可以从低位到高位逐级递加了
1 public class Solution { 2 public String addBinary(String a, String b) { 3 if( a == null || a.length() == 0) return b; 4 if( b == null || b.length() == 0) return a; 5 // assume that b.length is longer 6 if(b.length() < a.length()) return addBinary(b, a); 7 StringBuilder br = new StringBuilder(b).reverse(); 8 StringBuilder ar = new StringBuilder(a).reverse(); 9 int leap = 0; 10 for(int i = 0; i < br.length(); i++){ 11 int aNum = (i < ar.length())? (ar.charAt(i) == ‘0‘ ? 0 : 1) : 0; 12 int bNum = br.charAt(i) == ‘0‘ ? 0 : 1; 13 if(aNum + bNum + leap == 1){ 14 br.setCharAt(i, ‘1‘); 15 leap = 0; 16 }else if(aNum + bNum + leap == 2){ 17 br.setCharAt(i, ‘0‘); 18 leap = 1; 19 }else if(aNum + bNum + leap == 3){ 20 br.setCharAt(i, ‘1‘); 21 leap = 1; 22 } 23 } 24 if(leap == 1) br.append(‘1‘); 25 return br.reverse().toString(); 26 } 27 }
[leetcode]Add Binary,布布扣,bubuko.com
标签:style blog http color os io for re
原文地址:http://www.cnblogs.com/huntfor/p/3866502.html