655. Big Integer Addition【easy】

Given two non-negative integers num1 and num2 represented as string, return the sum of num1 and num2.

Notice

• The length of both num1 and num2 is < 5100.
• Both num1 and num2 contains only digits 0-9.
• Both num1 and num2 does not contain any leading zero.
• You must not use any built-in BigInteger library or convert the inputs to integer directly.

 

Example

Given num1 = "123", num2 = "45"
return "168"

解法一：

class Solution {  
public:  
    /** 
    * @param num1 a non-negative integers 
    * @param num2 a non-negative integers 
    * @return return sum of num1 and num2 
    */  
    string addStrings(string& num1, string& num2) {  
        // Write your code here  
        int add_bit = 0, i = 0;  
        char temp, result[6100] = { 0 };  
        const char *n1 = num1.c_str();  
        const char *n2 = num2.c_str();  
        int len1 = strlen(n1);  
        int len2 = strlen(n2);  
        while (len1 != 0 && len2 != 0) {  
            len1--; len2--;  
            result[i] = (add_bit + n2[len2] + n1[len1] - 2 * '0') % 10 + '0';  
            add_bit = (n2[len2] + n1[len1] + add_bit - '0' * 2) / 10;  
            i++;  
        }  
        if (len1 > len2) {  
            while (len1) {  
                len1--;  
                result[i] = (add_bit + n1[len1] - '0') % 10 + '0';  
                add_bit = (n1[len1] + add_bit - '0') / 10;  
                i++;  
            }  
        }  
        else {  
            while (len2) {  
                len2--;  
                result[i] = (add_bit + n2[len2] + add_bit- '0') % 10 + '0';  
                add_bit = (add_bit + n2[len2] - '0') / 10;  
                i++;  
            }  
        }  
        if (add_bit) {  
            result[i] = '1';  
            i++;  
        }  
        for (int j = 0; j < i / 2; j++)  
        {  
            temp = result[j];  
            result[j] = result[i - 1 - j];  
            result[i - 1 - j] = temp;  
        }  
        result[i] = '';  
        return result;  
    }  
}; 

写法太复杂

解法二：

class Solution {
public:
    /**
     * @param num1 a non-negative integers
     * @param num2 a non-negative integers
     * @return return sum of num1 and num2
     */
    string addStrings(string& num1, string& num2) {
        int m = num1.size();
        int n = num2.size();
        string result;
        int i = m - 1, j = n - 1;
        int carry = 0;
        while (i >= 0 || j >= 0) {
            int sum = carry;
            sum += (i >= 0) ? num1[i--] - '0' : 0;
            sum += (j >= 0) ? num2[j--] - '0' : 0;
            carry = sum / 10;
            sum %= 10;
            result += '0' + sum;
        }
        
        if (carry) {
            result += '1';
        }
        
        reverse(result.begin(), result.end());
        return result;
    }
};

参考@jiadaizhao 的代码

解法三：

public class Solution {
    /*
     * @param num1: a non-negative integers
     * @param num2: a non-negative integers
     * @return: return sum of num1 and num2
     */
    public String addStrings(String num1, String num2) {
        //start from adding the last digits of num1, num2:
        //if the current sum > 10, save 1 in `carry`,
        //add to the front of StringBuilder sb
        //... doing this till both indice less than 0
        
        int i = num1.length()-1, j = num2.length()-1, carry = 0, curSum = 0;
        StringBuilder sb = new StringBuilder();
        while (i >= 0 || j >= 0 || carry == 1) {
            //Integer.valueOf(String.valueOf(char)) is to remind me that the value of char is mapped to the decimal value in ascii 
            int curNum1 = i >= 0 ? Integer.valueOf(String.valueOf(num1.charAt(i))) : 0;
            int curNum2 = j >= 0 ? Integer.valueOf(String.valueOf(num2.charAt(j))) : 0;
            int sum = carry + curNum1 + curNum2;
            curSum = sum % 10; carry = sum/10;
            sb.insert(0, curSum);
            i--; j--;
        }
        return sb.toString();
    }
}

参考@linspiration 的代码

https://segmentfault.com/a/1190000012466338

解法四：

public class Solution {
    /**
     * @param num1 a non-negative integers
     * @param num2 a non-negative integers
     * @return return sum of num1 and num2
     */
    public String addStrings(String num1, String num2) {
        if (num1 == null || num1.length() == 0) {
            return num2;
        } 
        if (num2 == null || num2.length() == 0) {
            return num1;
        }
        int index1 = num1.length() - 1;
        int index2 = num2.length() - 1;
        int carry = 0;
        StringBuilder sb = new StringBuilder();
        while (index1 >= 0 || index2 >= 0) {
            if (index1 >= 0) {
                carry = carry + (num1.charAt(index1) - '0');
            }
            if (index2 >= 0) {
                carry = carry + (num2.charAt(index2) - '0');
            }
            sb.insert(0, carry % 10);
            carry = carry / 10;
            --index1;
            --index2;
        }
        if (carry > 0) {
            sb.insert(0, carry);
        }
        return sb.toString();
    }
}

参考@Microthinking 的代码

https://blog.happynavy.tk/lintcodes/big-integer-addition/