Given two non-negative integers num1 and num2 represented as string, return the sum of num1 and num2.
Note:
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.
分析：先将num1和num2倒置，然后从头到尾依次加，记得标记是否有进位sign，每次也要把进位的值加进去～如果num1加完了num2还没加完，就继续加num2的～反之同理～最后sign也要考虑有没有最高位1～所有都处理完后把result倒置～
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class Solution { public: string addStrings(string num1, string num2) { int len1 = num1.length(), len2 = num2.length(); for(int i = 0; i < len1 / 2; i++) swap(num1[i], num1[len1 - i - 1]); for(int i = 0; i < len2 / 2; i++) swap(num2[i], num2[len2 - i - 1]); string result = ""; int sign = 0, p = 0, q = 0; while(p < len1 && q < len2) { int t = (num1[p] - '0') + (num2[q] - '0') + sign; if(t >= 10) { t = t - 10; sign = 1; } else { sign = 0; } result += (char)(t + '0'); p++; q++; } while(p < len1) { int t = (num1[p] - '0') + sign; if(t >= 10) { t = t - 10; sign = 1; } else { sign = 0; } result += (char)(t + '0'); p++; } while(q < len2) { int t = (num2[q] - '0') + sign; if(t >= 10) { t = t - 10; sign = 1; } else { sign = 0; } result += (char)(t + '0'); q++; } if(sign == 1) result += '1'; int len = result.length(); for(int i = 0; i < len / 2; i++) { swap(result[i], result[len - 1 - i]); } return result; } }; |
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