LeetCode之67. Add Binary

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

The input strings are both non-empty and contains only characters 1 or 0.

Example 1:

Input: a = "11", b = "1"
Output: "100"

Example 2:

Input: a = "1010", b = "1011"
Output: "10101"

自己的代码：

Runtime: 1 ms, faster than 100.00% of Java online submissions for Add Binary.

Memory Usage: 37.2 MB, less than 12.48% of Java online submissions for Add Binary.

public static String addBinary(String a, String b) {
	if(a == null || a.length() == 0) return b;
	if(b == null || b.length() == 0) return a;

	if(a.length() < b.length()) { String temp = a; a = b; b = temp; }

	StringBuilder sb = new StringBuilder();
	int carry = 0; int cur = 0; int x = 0; 	int y = 0;

    // 阶段一
	for(int i=b.length()-1,j=a.length()-1;i>=0;i--,j--) {
		x = a.charAt(j) - '0'; 	y = b.charAt(i) - '0';
		cur = x + y + carry;

		carry = cur / 2; cur = cur % 2;
		sb.insert(0, cur);
	}

    // 阶段二
	for(int i=a.length()-b.length()-1;i>=0;i--) {
		x = a.charAt(i) - '0';
		cur = x + carry;

		carry = cur / 2; cur = cur % 2;
		sb.insert(0, cur);
	}

    // 阶段三
	if(carry != 0) sb.insert(0, carry);  // 最后不要忘了加上可能的进位
	
	return sb.toString();
}

          1） 整体处理分为三个阶段：

                 阶段一是用来处理长串与短串长度相同的那一截儿

                 阶段二是用来处理长串多出来的那一截儿

                 阶段三是处理一下最终可能存在的进位carry