LeetCode - Add Digits

LeetCode 的 Add Digits 題目如下:

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.


簡單的說就是給予一個整數,反覆的將每個位數值加總,直至數值只有一個位數為止。


最直覺得解法就是用迴圈下去處理,判斷數值是否大於等於 10,如果大於等於 10,則將每個位數的數值加總取代本來的數值。如果數值小於 10,則直接將數值回傳。

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public class Solution {
public int AddDigits(int num) {
var value = num;
while(value >= 10)
{
var originalValue = value;
var newValue = 0;
do
{
newValue += originalValue / 10;
originalValue = originalValue % 10;
}while(originalValue >= 10);

newValue += originalValue;

value = newValue;
}
return value;
}
}



但比較漂亮的解法是像下面這樣:

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