67. Add Binary

谁践踏了优雅 2022-06-13 05:58 124阅读 0赞

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

For example,  
a = `"11"`  
b = `"1"`

Return `"100"`.

问题描述：给定两个二进制字符串，返回这两个二进制字符串的和，和也是一个二进制字符串。

分析：循环这两个字符串，每次获得一个字符，转换成String，再转换成Int后相加，设置变量m,记录两数相加的进位。若两数相加再加进位m大于或等于2，则进位m=1,当前位为两数相加再加进位m%2。解题思路和[66. Plus One][] 类似。注意最后相加结果result反转后才是最终结果。

public class Solution {
        public String addBinary(String a, String b) {
    		int m = 0;
    		String result = "";
    		int i = a.length() - 1;
    		int j = b.length() - 1;
    		int a1;
    		int b1;
    		for (; i >= 0 && j >= 0; i--, j--) {
    			a1 = Integer.parseInt(String.valueOf(a.charAt(i)));
    			b1 = Integer.parseInt(String.valueOf(b.charAt(j)));
    			if ((a1 + b1 + m) >= 2) {
    				result = result + (a1 + b1 + m) % 2;
    				m = 1;
    			} else {
    				result = result + (a1 + b1 + m);
    				m = 0;
    			}
    		}
    		if (i >= 0) {
    			for (; i >= 0; i--) {
    				a1 = Integer.parseInt(String.valueOf(a.charAt(i)));
    				if (a1 + m == 2) {
    					result = result + 0;
    					m = 1;
    					if (i == 0) {
    						result += 1;
    					}
    				} else {
    					result = result + (a1 + m);
    					m = 0;
    
    				}
    			}
    		} else if (j >= 0) {
    			for (; j >= 0; j--) {
    				b1 = Integer.parseInt(String.valueOf(b.charAt(j)));
    				if (b1 + m == 2) {
    					result = result + 0;
    					m = 1;
    					if (j == 0) {
    						result += 1;
    					}
    				} else {
    					result = result + (b1 + m);
    					m = 0;
    				}
    			}
    		} else {
    			if (m == 1) {
    				result += 1;
    			}
    		}
    		String newResult = "";
    		for (int k = result.length() - 1; k >= 0; k--)
    			newResult += result.charAt(k);
    		return newResult;
        }
    }

[66. Plus One]: http://blog.csdn.net/wust_lh/article/details/73527572