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 = , i = ;
         char temp, result[] = {  };
         const char *n1 = num1.c_str();
         const char *n2 = num2.c_str();
         int len1 = strlen(n1);
         int len2 = strlen(n2);
         while (len1 !=  && len2 != ) {
             len1--; len2--;
             result[i] = (add_bit + n2[len2] + n1[len1] -  * '') %  + '';
             add_bit = (n2[len2] + n1[len1] + add_bit - '' * ) / ;
             i++;
         }
         if (len1 > len2) {
             while (len1) {
                 len1--;
                 result[i] = (add_bit + n1[len1] - '') %  + '';
                 add_bit = (n1[len1] + add_bit - '') / ;
                 i++;
             }
         }
         else {
             while (len2) {
                 len2--;
                 result[i] = (add_bit + n2[len2] + add_bit- '') %  + '';
                 add_bit = (add_bit + n2[len2] - '') / ;
                 i++;
             }
         }
         if (add_bit) {
             result[i] = '';
             i++;
         }
         for (int j = ; j < i / ; j++)
         {
             temp = result[j];
             result[j] = result[i -  - j];
             result[i -  - j] = temp;
         }
         result[i] = '\0';
         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 - , j = n - ;
         int carry = ;
         while (i >=  || j >= ) {
             int sum = carry;
             sum += (i >= ) ? num1[i--] - '' : ;
             sum += (j >= ) ? num2[j--] - '' : ;
             carry = sum / ;
             sum %= ;
             result += '' + sum;
         }

         if (carry) {
             result += '';
         }

         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/