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?
Hint:
数学题给跪了。
如果可以用循环，最好的情况是读一遍数字，O(1)时间复杂度。
贪心，每读一位数都做求Digit Root的操作，最后的结果是正确的。
1 /** 2 * @param {number} num 3 * @return {number} 4 */ 5 var addDigits = function(num) { 6 var str = num.toString(), res = 0, tmp1, tmp2; 7 for(var i = 0; i < str.length; i++){ 8 res = parseInt(str[i]) + res; 9 if(res >= 10){ 10 tmp1 = parseInt(res / 10); 11 tmp2 = res % 10; 12 res = tmp1 + tmp2; 13 } 14 } 15 return res; 16 };
最佳答案参照wiki ： https://en.wikipedia.org/wiki/Digital_root
好几种写法都可以，这里列了2种。
1 /** 2 * @param {number} num 3 * @return {number} 4 */ 5 var addDigits = function(num) { 6 return num === 0 ? 0 : num - 9 * Math.floor((num - 1) / 9); 7 };
1 /** 2 * @param {number} num 3 * @return {number} 4 */ 5 var addDigits = function(num) { 6 return 1 + (num - 1) % 9; 7 };
[LeetCode][JavaScript]Add Digits
原文：http://www.cnblogs.com/Liok3187/p/4735356.html