LeetCode 202. Happy Number _leetcode

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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.
【LeetCode 202. Happy Number】Example: 19 is a happy number

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

Credits:
Special thanks to @mithmatt and @ts for adding this problem and creating all test cases.
题意：给定一个整数，判断该数是否为Happy Number。
Happy Number的定义：计算整数的各个位上数字的平方和，重复该过程，如果出现和为1，则说明该数是Happy Number。如果在一个数字圈中无限循环，且该数字圈不包括1，则说明该数不是Happy Number。

方法一：开始理解错题意，想成了如果一个数不是Happy Number，则会出现平方和一直为某一个值的无限循环。提交不通过后又读一遍题，发现是如果一个数不是Happy Number，则平方和会从某处开始一直循环（想想也是这个道理）。
利用Set不能存放重复元素的特性，将每次的平方和都存入set，如果某次的平方和sum在set中已经出现过，则说明已经进入循环，该数字不是一个Happy Number。
beats 18.50 % of java submissions.

public boolean isHappy(int n) { int sum = 0, temp; Set< Integer> set = new HashSet< > (); while(!set.contains(n)){ set.add(n); while(n != 0){ temp = n % 10; sum += temp * temp; n = n / 10; } if(sum == 1) return true; n = sum; sum = 0; } return false; }

方法二：将每次计算的平方和存入一个链表，利用快慢指针，如果快指针跟慢指针重逢，则说明存在一个环，该数不是Happy Number。
beats 78.05 % of java submissions.

public boolean isHappy(int n) { int sum = 0, temp; Node head = new Node(n); Node fast = head, slow = head, test = head; while(slow == test || fast.val != slow.val){//slow == test用于排除fast和slow都指向第一个结点的初试情况 while(n != 0){ temp = n % 10; sum += temp * temp; n = n / 10; } if(sum == 1) return true; n = sum; sum = 0; head.next = new Node(n); head = head.next; if(fast.next.next != null){ fast = fast.next.next; slow = slow.next; } } return false; }class Node{ int val; Node next; public Node(int val){ this.val = val; } }

方法三：只是与方法二的循环结束条件不同

public boolean isHappy(int n) { int sum = 0, temp; Node head = new Node(n); Node fast = head, slow = head, test = head; ; while(n != 1){ while(n != 0){ temp = n % 10; sum += temp * temp; n = n / 10; } n = sum; sum = 0; head.next = new Node(n); head = head.next; if(fast.next.next != null){ fast = fast.next.next; slow = slow.next; } if(slow != test & & fast.val == slow.val) return false; } return true; }

方法四：利用函数来取代快慢指针，参考：https://www.jianshu.com/p/f7b632e31d5f
beats 78.05 % of java submissions.

public boolean isHappy(int n){ int sum = 0; int fast = n, slow = n; while(sum != 1){ slow = getSum(slow); if(slow == 1) return true; fast = getSum(getSum(fast)); if(fast == slow) return false; sum = slow; } return true; } public int getSum(int n){ int temp, sum = 0; while(n != 0){ temp = n % 10; sum += temp * temp; n /= 10; } return sum; }