leetcode 202 Happy Number

202. Happy Number

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

Tags:Hash Table; Math

解題思路：直接根據題目所示步驟編程，重複計算平方和，得到一系列數字，當某個數重複出現時（可看成這些數組成的鏈 存在圈），則將無休止地繼續計算平方和，因此不是Happy Number。

錯誤範例：

class Solution {
private:
    int squaredAdd(int num){
        int sum = 0, digit;
        while(num){
            digit = num % 10;
            sum += digit * digit;
            num /= 10;
        }
        return sum;
    }
public:
    bool isHappy(int n) {
        int number = n;
        while(1){
            number = squaredAdd(number);
            if(number == 1)  return true;
            if(number == n)  return false; //此處錯誤！
        }
    }
};

錯誤原因：數鏈存在圈，但交匯點不一定在數字n處（不一定在n處開始進入圈）。以n＝2爲例，重複計算平方和得到數鏈：2，4，16，37，58，89，145，42，20，4，16... 將數鏈寫成環形可發現在數字4進入圈，而不是2，所以此方法錯誤。

正確方法：可參考檢測鏈表中是否有圈的方法，使用兩個指針，快指針和慢指針。快指針每次走兩步，慢指針每次走一步，若存在圈，則快指針會追上慢指針，二者終將相等。

class Solution {
private:
    int squaredAdd(int num){
        int sum = 0, digit;
        while(num){
            digit = num % 10;
            sum += digit * digit;
            num /= 10;
        }
        return sum;
    }
public:
    bool isHappy(int n) {
        int slow = n, fast = n;
        while(1){
            slow = squaredAdd(slow);
            fast = squaredAdd(fast);
            fast = squaredAdd(fast);
            if(fast == 1)   return true;    //必須先判斷fast是否等於1，否則會出錯
            if(fast == slow)    return false;    //eg. n爲1時應返回true，若先判斷fast是否等於slow，則會返回false，錯誤！
        }
    }
};