67. 二进制求和 - 力扣（LeetCode）

两个题目方法完全一样

用两个数据的末尾位相加，从末尾位开始逐位相加，记录进位；

class Solution {
public:string addBinary(string a, string b) {int end1 = a.size() - 1;int end2 = b.size() - 1;string num;//最终的结果int next = 0;//进位    while(end1 >= 0 || end2 >= 0){int val1 = 0, val2 = 0;if(end1 >= 0)val1 = a[end1] - '0';if(end2 >= 0)val2 = b[end2] - '0';int sum = val1 + val2 + next;if(sum > 1){sum -= 2;next = 1;}else{next = 0;}num.push_back(sum + '0');end1--;end2--;}if(next == 1){num.push_back('1');}reverse(num.begin(),num.end());return num;}
};

reverse是stl里面的算法，就是逆置；

+=都是头插；

方法是先用个位相加，然后有进位的标记；

其中首先他是字符所以减去'0',ascll码相减就会变成能加减的数值，最后还要转变回来，所以+'0'

415. 字符串相加 - 力扣（LeetCode）

class Solution {
public:string addStrings(string num1, string num2) {int end1 = num1.size() - 1;int end2 = num2.size() - 1;string retstr;int next = 0;//进位while(end1 >= 0 || end2 >= 0){int val1 = 0, val2 = 0;if(end1 >= 0)val1 = num1[end1] - '0';//字符串相减if(end2 >= 0)val2 = num2[end2] - '0';int ret = val1 + val2 + next;if(ret > 9){ret -= 10;next = 1;}else{next = 0;}retstr += (ret + '0');--end1;--end2;}if(next == 1){retstr += "1";}reverse(retstr.begin(), retstr.end());return retstr;}
};