Leetcode67-二进制求和

给你两个二进制字符串，返回它们的和（用二进制表示）。输入为非空字符串且只包含数字 1 和 0。

示例 1:

输入: a = "11", b = "1"
输出: "100"

示例 2:

输入: a = "1010", b = "1011"
输出: "10101"

 

本题的要点在于注意进位以及未处理完的那个字符串。

 1 // Leetcode 67
 2 // Math, string
 3 var addBinary = function(a, b) {
 4     if (a.length == 0 && b.length == 0) {
 5         return "";
 6     }
 7 
 8     if (a.length == 0 && b.length != 0) {
 9         return b;
10     }
11 
12     if (a.length != 0 && b.length == 0) {
13         return a;
14     }
15 
16     PointerA = a.length - 1; PointerB = b.length - 1; carry = 0; res = "";
17     while (PointerA >= 0 && PointerB >= 0) {
18         tempSum = parseInt(a[PointerA]) + parseInt(b[PointerB]) + carry;
19         if (tempSum >= 2) {
20             carry = 1;
21             res += (tempSum % 2).toString();
22         } else {
23             carry = 0;
24             res += (tempSum % 2).toString();
25         }
26 
27         PointerA--;
28         PointerB--;
29     }
30     
       //如果A没有处理完，则继续处理A
31     if (PointerA >= 0) {
32         if (carry == 0) {
33             for (let i = PointerA; i >= 0; i--) {
34                 res += a[i];
35             }
36         } else {
37             carry = 1;
38             for (let i = PointerA; i >= 0; i--) {
39                 tempSum = parseInt(a[i]) + carry;
40                 if (tempSum >= 2) {
41                     carry = 1;
42                     res += (tempSum % 2).toString();
43                 } else {
44                     carry = 0;
45                     res += (tempSum % 2).toString();
46                 }
47             }
48         }
49     }

50     //如果B没有处理完，则继续处理B
51     if (PointerB >= 0) {
52         if (carry == 0) {
53             for (let i = PointerB; i >= 0; i--) {
54                 res += b[i];
55             }
56         } else {
57             carry = 1;
58             for (let i = PointerB; i >= 0; i--) {
59                 tempSum = parseInt(b[i]) + carry;
60                 if (tempSum >= 2) {
61                     carry = 1;
62                     res += (tempSum % 2).toString();
63                 } else {
64                     carry = 0;
65                     res += (tempSum % 2).toString();
66                 }
67             }
68         }
69     }
70 
71     if (carry == 1) {
72         res += "1"; //注意最后的进位
73     }
74 
75     return res.split("").reverse().join("");
76 };