标签：add color contains recursion ber 公式 you turn uil
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?
var addDigits = function(num) { var res = num; while(res >= 10){ num = res; res = 0; while(num>0){ res = res + num % 10; num = parseInt(num / 10); } } return res; };
题目后续：写成不用循环的方法，查了一下百度发现还有数学公式来着，就可以不用循环了0_0。
公式是(num-1)%9+1
Given two non-negative integers num1 and num2 represented as string, return the sum of num1 and num2.
Note:
num1 and num2 is < 5100.num1 and num2 contains only digits 0-9.num1 and num2 does not contain any leading zero.var addStrings = function(num1, num2) { if(num1.length < num2.length) { var _swap = num1; num1 = num2; num2 = _swap; } var i = num1.length -1; var j = num2.length -1; var res = []; var cur=0,cin=0; while(j>=0){ cur = (cin + parseInt(num1[i]) + parseInt(num2[j])) % 10; cin = parseInt((cin + parseInt(num1[i]) + parseInt(num2[j])) / 10); res.unshift(cur); j--; i--; } while(i>=0){ if(cin>0){ cur = (cin + parseInt(num1[i])) % 10; cin = parseInt( (cin + parseInt(num1[i]) ) / 10); res.unshift(cur); } else { res.unshift(parseInt(num1[i])); } i--; } if(cin>=1){ res.unshift("1"); } return res.join(‘‘); };
leetcode练习：258. Add Digits & 415. Add Strings
标签：add color contains recursion ber 公式 you turn uil
原文地址：http://www.cnblogs.com/rimochiko/p/7725253.html
