力扣67 二进制求和

news2024/12/22 17:25:27

文章目录

1. 题目链接
2. 题目代码
3.感受

1. 题目链接

二进制求和

2. 题目代码

class Solution {
public:
    string addBinary(string a, string b) {
        vector<int> stringA;
        vector<int> stringB;
        int lengthOfA = a.length();
        int lengthOfB = b.length();
        for(int subscriptOfA = 0; subscriptOfA < lengthOfA; subscriptOfA ++){
            stringA.push_back(a[subscriptOfA] - '0');
        }
        for(int subscriptOfB = 0; subscriptOfB < lengthOfB; subscriptOfB ++){
            stringB.push_back(b[subscriptOfB] - '0');
        }
        if(lengthOfA > lengthOfB){
            stringB.insert(stringB.begin(), lengthOfA - lengthOfB , 0);
        }
        if(lengthOfB > lengthOfA){
            stringA.insert(stringA.begin(), lengthOfB - lengthOfA , 0);
        }
        reverse(stringA.begin(), stringA.end());
        reverse(stringB.begin(), stringB.end());
        vector<int> result;
        int carry = 0;
        for(int subscript = 0; subscript < stringA.size(); subscript ++){
            result.push_back((stringA[subscript] + stringB[subscript] + carry) % 2);
            if(stringA[subscript] + stringB[subscript] + carry >= 2){
                carry = 1;
            }else{
                carry = 0;
            }
        }
        if(carry == 1){
            result.push_back(1);
        }
        string finallResult;
        for(int subscript = result.size() - 1; subscript >= 0; subscript --){
            finallResult += result[subscript] + '0';
        }
        return finallResult;
    }
};