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Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.
For example:
Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.
Follow up:
Could you do it without any loop/recursion in O(1) runtime?
思路：如果不对复杂度有要求，通过常规算法，不断对10求余和除10，一步一步用for循环相加就能得出结果。但题目中要求不用循环，看了几个hint后，感觉有规律，但仍无法找出具体公式，后来看了wiki百科中的公式后，发现就是1-9不断的循环，问题变得很简单，加起来的数可以归结为一个公式，
之后的代码就很简单了。
我的代码：
class Solution { public: int addDigits(int num) { int s = 1 + (num-1)%9; return s; } };
参考代码：
class Solution { public: int addDigits(int num) { switch(random()%5+1){ case 1: return addDigits01(num); case 2: return addDigits02(num); case 3: return addDigits03(num); case 4: return addDigits04(num); default: return addDigits05(num); } } //regualr way int addDigits01(int num) { while(num > 9) { int sum; for(sum=0; num > 0; sum += num%10 , num/=10); num = sum; } return num; } //This solution looks is very tricky, but acutally it is easy to understand. //it just keep adding the last digital until the num < 10 int addDigits02(int num) { while(num > 9) { num = num / 10 + num % 10; } return num; } // Let‘s observe the pattern // 1 1 // 2 2 // ... ... // 8 8 // 9 9 // 10 1 // 11 2 // 12 3 // ... ... // 17 8 // 18 9 // 19 1 // 20 2 // ... ... // It looks most of number just simply %9 is the answer, // but there are some edge cases. // 9%9=0 but we need 9. // 18%9=0 but we need 9 // so we can find the solution is: // 1) num <=9, return num // 2) num > 9, reutrn num%9 if num%9>0 // return 9 if num%9 ==0 int addDigits03(int num) { return num >9 ? ((num %9)==0 ? 9:num%9) : num; } //But actually, we can use (num-1)%9 + 1 to make all cases right. int addDigits04(int num){ return (num - 1) % 9 + 1; } //This solution is similar with pervious solution. int addDigits05(int num){ return num - 9 * ((num - 1)/9); } };
总结：参考代码完整得展现了从常规算法，到最简算法的历程，在两三个条件判断下，学习参考代码中的写法：
return num >9 ? ((num %9)==0 ? 9:num%9) : num;
而对同一问题有多种算法时，用switch＋random（）的组合非常好玩，哈哈：
switch(random()%5+1)
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原文地址：http://www.cnblogs.com/torrescx/p/5205813.html
