标签：btn code container indent 计算公式 pad new 复杂 padding
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?
题目让求各个数位的和，直到和小于10为止，
直观的想法就是直接相加，先想一下如何求各数位的和
while(num != 0)
{
tmp += num % 10;
num /= 10;
}class Solution {
public int addDigits(int num) {
while(num / 10 > 0)
{ int sum = 0; ////注意定义变量的位置
while(num != 0)
{
sum += num % 10;
num /= 10;
}
num = sum;
}
return num;
}
}1 1
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20 2class Solution {
public:
int addDigits(int num) {
return (num - 1) % 9 + 1; //周期第一位为1，若为2 则 + 2
}
};标签：btn code container indent 计算公式 pad new 复杂 padding
原文地址：http://www.cnblogs.com/wxshi/p/7598531.html
