[LC]202. happy numbers

一、问题描述

Write an algorithm to determine if a number is "happy".

A happy number is a number defined by the following process: Starting with any positive integer, replace the number by the sum of the squares of its digits, and repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1. Those numbers for which this process ends in 1 are happy numbers.

Example: 19 is a happy number

• 12 + 92 = 82
• 82 + 22 = 68
• 62 + 82 = 100
• 12 + 02 + 02 = 1

二、我的思路

用一个ArrayList存储中间变量，每次判断一下当前中间变量是否出现过来判断是否endless。如果中间变量是1则是happy number。

class Solution {    public boolean isHappy(int n) {        List<Integer> appeared = new ArrayList<Integer>();        int cur_n;        while(n != 1 && (!appeared.contains(n))){            appeared.add(n);            cur_n = 0;            while(n > 0){                cur_n += (n%10) * (n%10);                n = n/10;            }            n = cur_n;        }        return n == 1;    }}

反思：

1. 用ArrayList存储是否是最好的？ArrayList是怎么实现contains函数的？

A：不是最好的，参见这篇文章http://blog.csdn.net/fenglibing/article/details/9021201。其中提到慎用ArrayList的contains方法。因为该方法的实现调用了很多indexOf方法，也就是说最坏情况下可能要查找多次List这么长的内容。而HashSet用的HashTable，所以速度会快很多

2. n%10可以先用变量存一下，就不用计算两次了。

修正后：

class Solution {    public boolean isHappy(int n) {        Set<Integer> appeared = new HashSet<Integer>();        int cur_n, remained;        while(n != 1 && (!appeared.contains(n))){            appeared.add(n);            cur_n = 0;            while(n > 0){                remained = (n%10);                cur_n += remained * remained;                n = n/10;            }            n = cur_n;        }        return n == 1;    }}

再看下高票解法，虽然思路一样，但是实现就能看出差距吖！他直接拿HashSet的add方法的返回参数做跳出循坏的条件了

public boolean isHappy(int n) {    Set<Integer> inLoop = new HashSet<Integer>();    int squareSum,remain;while (inLoop.add(n)) {squareSum = 0;while (n > 0) {    remain = n%10;squareSum += remain*remain;n /= 10;}if (squareSum == 1)return true;elsen = squareSum;}return false;}

三、淫奇技巧

1. 空间上O(1)解法，思路借鉴了判断单链表是否成环：一个指针走一步，另一个走两步，当两个人当值相等了（且不是1）就false.

public class Solution {    public boolean isHappy(int n) {        int x = n;        int y = n;        while(x>1){            x = cal(x) ;            if(x==1) return true ;            y = cal(cal(y));            if(y==1) return true ;            if(x==y) return false;        }        return true ;    }    public int cal(int n){        int x = n;        int s = 0;        while(x>0){            s = s+(x%10)*(x%10);            x = x/10;        }        return s ;    }}