问题描述
Given two binary strings, return their sum (also a binary string).
For example,
a = "11"
b = "1"
Return "100".
算法
代码一
1 public String addBinary(String a, String b) { 2 int i = 0, length1 = a.length(), length2 = b.length(); 3 HashMap<String, String> map = new HashMap<String, String>(); 4 map.put("000", "00"); 5 map.put("001", "01"); 6 map.put("010", "01"); 7 map.put("011", "10"); 8 map.put("100", "01"); 9 map.put("101", "10"); 10 map.put("110", "10"); 11 map.put("111", "11"); 12 String s = "0", result = "", value; 13 while (i < length1 && i < length2) { 14 s += a.charAt(length1 - 1 - i); 15 s += b.charAt(length2 - 1 - i); 16 value = map.get(s); 17 result += value.charAt(1); 18 s = value.substring(0, 1); 19 i++; 20 } 21 if (i >= length1) { 22 while (i < length2) { 23 s += b.charAt(length2 - 1 - i); 24 s += "0"; 25 value = map.get(s); 26 result += value.charAt(1); 27 s = value.substring(0, 1); 28 i++; 29 } 30 } else { 31 while (i < length1) { 32 s += a.charAt(length1 - 1 - i); 33 s += "0"; 34 value = map.get(s); 35 result += value.charAt(1); 36 s = value.substring(0, 1); 37 i++; 38 } 39 } 40 if(s.equals("1")) 41 result+=s; 42 return new StringBuilder(result).reverse().toString(); 43 }
代码二
1 public String addBinary1(String a, String b) { 2 StringBuilder sb = new StringBuilder(); 3 int i = a.length() - 1, j = b.length() - 1, carry = 0; 4 while (i >= 0 || j >= 0) { 5 int sum = carry; 6 if (j >= 0) 7 sum += b.charAt(j--) - '0'; 8 if (i >= 0) 9 sum += a.charAt(i--) - '0'; 10 sb.append(sum % 2); 11 carry = sum / 2; 12 } 13 if (carry != 0) 14 sb.append(carry); 15 return sb.reverse().toString(); 16 17 }
注意事项：
（1）String s="sdhfowf";则s.charAt(0)=='s'。
（2）String变量没有反转函数，StringBuilder有反转函数。