LeetCode题解:Add Binary
2015-09-08 08:27
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题目链接:
Add Binary题目描述:
Given two binary strings, return their sum (also a binary string).For example,
a = “11”
b = “1”
Return “100”.
题目解释:
给定两个01串(字符串内的字符全是0或者1),返回这连个串相加的结果。例如: 串a=”11”、串b=”1”,那么这两个串相加的结果等于”100”。
解题方案:
这个题目实际上是一个二进制加法的实现,我们用字符串存储二进制数,然后相加,满2要进位,我们迭代处理字符串每一位的相加并加上之前位相加时的进位即可,但是这里要注意一个问题:我们在做算数运算时,数的高位是放在数位的左侧的,比如我们正常的数字110,这个二进制数的最高位是1而不是0,也就是说按正常来我们的加法是这样子的:——— 1 1 0
—————1 结果为===> 111,
而在字符串中我们的二进制数就是按高位放串左端的方式,但这给我们实际操作带来了麻烦,我们得将两个串右对齐以后才能进行假发操作,太不方便,为了简便迭代过程,我先将字符串翻转了一下,这样子以后,就是左对齐了。
当然,这里还有一点,我没有重新开辟一个存储空间,我在a、b两个串中比较长的那个串上进行的相应操作,所以我的代码显得比较冗余,下面我们来看一下具体实现的代码:
class Solution { public: string addBinary(string a, string b) { reverse(a.begin(), a.end()); reverse(b.begin(), b.end()); //int size = (a.length() > b.length())? a.length():b.length(); int add = 0; //string rs; //当串a的长度大于串b上时,将串a作为基础串来进行相应的操作 if(a.length() >= b.length()) { add = 0; for (int i = 0; i < a.length(); ++i) { if(i >= b.size()) { if((a[i] + add - 48) > 1) { a[i] += (add-2); add = 1; } else { a[i] += add; add = 0; } } else { if((a[i] + b[i] + add - 96) > 1) { a[i] += (b[i] + add - 48 - 2); add = 1; } else { a[i] += b[i] + add - 48; add = 0; } } } if(add) { a.push_back('1'); } reverse(a.begin(), a.end()); return a; } //当串a的长度小于串b上时,将串b作为基础串来进行相应的操作 if(a.length() < b.length()) { add = 0; for (int i = 0; i < b.length(); ++i) { if(i >= a.size()) { if((b[i] + add - 48) > 1) { b[i] += (add-2); add = 1; } else { b[i] += add; add = 0; } } else { if((a[i] + b[i] + add - 96) > 1) { b[i] += (a[i] + add - 48 - 2); add = 1; } else { b[i] += a[i] + add - 48; add = 0; } } } if(add) { b.push_back('1'); } reverse(b.begin(), b.end()); return b; } // return a; } };
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