[LeetCode] Add Digits

标签：c++   leetcode   

Add Digits

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?

解题思路：

解法一：

模拟法，不断将数字拆分然后相加，直到只有一位数。

class Solution {
public:
    int addDigits(int num) {
        while(num/10>0){
            int sum = 0;
            while(num!=0){
                sum += num%10;
                num = num/10;
            }
            num = sum;
        }
        return num;
    }
};

题目要求没有递归，没有循环，且时间复杂度为O(1)。那就找规律。

~input: 0 1 2 3 4 ...
output: 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 ....

有两个公式：

d(n) = num%9 == 0 ? (num==0? 0 : 9) : num%9

d(n) = 1 + (n-1) % 9

因此代码如下：

class Solution {
public:
    int addDigits(int num) {
        return 1 + (num-1)%9;
    }
};

class Solution {
public:
    int addDigits(int num) {
        return num%9 == 0? (num==0?0:9) : num%9;
    }
};

[LeetCode] Add Digits

标签：c++   leetcode   

原文地址：http://blog.csdn.net/kangrydotnet/article/details/47721139