258.Add Digits
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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? */int a[101];int addDigits(int num) { int i; while(num>=10) { num = getArray(num);//直到num小于10才返回 } return num;}int getArray(int total){ int i,count = 0, res = 0; for(i = 0; total != 0 ; i++)//将数字分配到数组中 { a[i] = total % 10; total = total / 10; count++; } for(i = 0 ; i < count ; i++)//累加起来 res = res + a[i]; return res;}
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