36进制加法

C 解法, 执行用时: 67ms, 内存消耗: 6280KB, 提交时间: 2022-06-26

/**
 * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
 *
 * 
 * @param A string字符串 
 * @param B string字符串 
 * @return string字符串
 *
 * C语言声明定义全局变量请加上static，防止重复定义
 */
void reverse(char *s) {
    int sLen = strlen(s);
    int halfLen = sLen / 2;
    char tmp;
    for (int i = 0; i < halfLen; i++) {
        tmp = s[i];
        s[i] = s[sLen - 1 -i];
        s[sLen - 1 - i] = tmp;
    }
}

int max(int a, int b) {
    return a > b ? a : b;
}

int charToInt(char a) {
    if (a >= '0' && a <= '9') {
        return a - '0';
    } else if (a >= 'a' && a <= 'z') {
        return a - 'a' + 10;
    } else {
        return 0;
    }
}

char intToChar(int a) {
    int m = a % 36;
    
    if (m < 10) {
        return m + '0';
    } else {
        return (m - 10) + 'a';
    }
}

char* thirtysixAdd(char* A, char* B ) {
    reverse(A);
    reverse(B);
    int ALen = strlen(A), BLen = strlen(B);
    char *result = (char *)malloc(sizeof(char) * (max(ALen, BLen) + 2));
    memset(result, 0, sizeof(char) * (max(ALen, BLen) + 2));
    
    int i = 0, carry = 0, v;
    while(i < ALen || i < BLen || carry > 0) {
        v = carry;
        if (i < ALen) {
            v += charToInt(A[i]);
        }
        if (i < BLen) {
            v += charToInt(B[i]);
        }
        
        carry = v / 36;
        v = v % 36;
        result[i++] = intToChar(v);
    }
    
    reverse(A);
    reverse(B);
    reverse(result);
    
    return result;
}

C++ 解法, 执行用时: 67ms, 内存消耗: 12060KB, 提交时间: 2022-06-25

class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param A string字符串 
     * @param B string字符串 
     * @return string字符串
     */
    string add(string &s1, string &s2)
    {
        string res = "";
        int carry = 0;
        for (int i = 0; i < s1.length(); i++)
        {
            int a, b;
            if (s1[i] >= '0' && s1[i] <= '9')
            {
                a = s1[i] - '0';
            }
            else
            {
                a = s1[i] - 87;
            }
            if (s2[i] >= '0' && s2[i] <= '9')
            {
                b = s2[i] - '0';
            }
            else
            {
                b = s2[i] - 87;
            }
            int t = a + b + carry;
            carry = t / 36;
            t = t % 36;
            if (t < 10)
            {
                res.append(1, '0' + t);
            }
            else
            {
                res.append(1, 'a' + t - 10);
            }
        }
        if(carry>0)
        {
            if (carry < 10)
            {
                res.append(1, '0' + carry);
            }
            else
            {
                res.append(1, 'a' + carry - 10);
            }
        }
        return res;
    }

    string thirtysixAdd(string s1, string s2)
    {
        // write code here
        reverse(s1.begin(), s1.end());
        reverse(s2.begin(), s2.end());
        while (s1.length() < s2.length())
        {
            s1.append(1, '0');
        }
        while (s2.length() < s1.length())
        {
            s2.append(1, '0');
        }
        string res=add(s1,s2);
        reverse(res.begin(),res.end());
        return res;
    }
};

C++ 解法, 执行用时: 68ms, 内存消耗: 13036KB, 提交时间: 2022-06-10

class Solution {
public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     * 
     * @param A string字符串 
     * @param B string字符串 
     * @return string字符串
     */
	int CharToNum(char ch) {
		if(isdigit(ch)) {
			return ch - '0';
		} else {
			return ch - 'a' + 10;
		}
	}

	char NumToChar(int num) {
		if(num < 10) { // 0 ~ 9
			return num + '0';
		} else { // 10 ~ 35
			return num - 10  + 'a';
		}
	}

	string thirtysixAdd(string A, string B) {
		int sizeA = A.size();
		int sizeB = B.size();

		vector<char> resVec(sizeA + sizeB);

		int index = resVec.size() - 1;
		int indexA = sizeA - 1, indexB = sizeB - 1;

		int add = 0;
		while(indexA >= 0 || indexB >= 0 || add > 0) {
			int numA = indexA >= 0 ? CharToNum(A[indexA]) : 0;
			int numB = indexB >= 0 ? CharToNum(B[indexB]) : 0;

			int val = numA + numB + add;
			add = val / 36;
			val = val % 36;

			resVec[index] = NumToChar(val);

			index--;
			indexA--;
			indexB--;
		}

		index = 0;
		while(resVec[index] == 0) {
			index++;
		}
		string res;
		for(int i = index; i < resVec.size(); i++) {
			res.push_back(resVec[i]);
		}

		return res;
	}
};

C++ 解法, 执行用时: 69ms, 内存消耗: 12140KB, 提交时间: 2022-05-23

class Solution {
  public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param A string字符串
     * @param B string字符串
     * @return string字符串
     */
    string num = "0123456789abcdefghijklmnopqrstuvwxyz";
    string thirtysixAdd(string A, string B) {
        string res;
        int i = A.size() - 1;
        int j = B.size() - 1;
        int temp = 0;
        while (i >= 0 && j >= 0) {
            int value_a = num.find(A[i]);
            int value_b = num.find(B[j]);
            int sum =  value_a + value_b + temp;
            if (sum >= 36) {
                temp = 1;
                res += num[sum % 36];
            } else {
                temp = 0;
                res += num[sum];
            }
            i--, j--;
        }
        while (i >= 0) {
            int sum = num.find(A[i]) + temp;
            if (sum >= 36) {
                temp = 1;
                res += num[sum % 36];
            } else {
                temp = 0;
                res += num[sum];
            }
            i--;
        }

        while (j >= 0) {
            int sum  = num.find(B[j]) + temp;
            if (sum >= 36) {
                temp = 1;
                res += num[sum % 36];
            } else {
                temp = 0;
                res += num[sum];
            }
            j--;
        }
        if (temp) {
            res += '1';
        }
        reverse(res.begin(), res.end());
        return res;
    }
};

C++ 解法, 执行用时: 69ms, 内存消耗: 12140KB, 提交时间: 2022-05-04

class Solution {
  public:
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param A string字符串
     * @param B string字符串
     * @return string字符串
     */
    string num = "0123456789abcdefghijklmnopqrstuvwxyz";
    string thirtysixAdd(string A, string B) {
        // write code here
        // 倒过来相加
        string res;
        int i = A.size() - 1;
        int j = B.size() - 1;
        int index1, index2, sum, carry = 0;
        while (i >= 0 && j >= 0) {
            index1 = num.find(A[i]);
            index2 = num.find(B[j]);
            sum =  index1 + index2 + carry;
            if (sum >= 36) {
                carry = 1;
                res += num[sum % 36];
            } else {
                carry = 0;
                res += num[sum];
            }
            i--, j--;
        }
        //A有剩余
        while (i >= 0) {
            sum = num.find(A[i]) + carry;
            if (sum >= 36) {
                carry = 1;
                res += num[sum % 36];
            } else {
                carry = 0;
                res += num[sum];
            }
            i--;
        }
        //B有剩余
        while (j >= 0) {
            sum  = num.find(B[j]) + carry;
            if (sum >= 36) {
                carry = 1;
                res += num[sum % 36];
            } else {
                carry = 0;
                res += num[sum];
            }
            j--;
        }
        if (carry) {
            res += '1';
        }
        reverse(res.begin(), res.end());
        return res;
    }
};

详情