LeetCode

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

做法比较蠢，循环了一百遍让他去跑，判断每次的结果是不是happy number，虽然AC了，但是不知道会不会有数据卡掉100。时间复杂度O(<3200)，空间复杂度O(1)

评论区的做法比较合理，不过心里有个疑问，如果slow和fast永远不等呢？如果有大神能证明这个算法的绝对正确性千万记得告诉窝！时间复杂度O(不会算orz)，空间复杂度O(1)

class Solution {public:    bool isHappy(int n) {        int flag = false;        int cntsum = 0;        while (cntsum < 100) {            cntsum++;            int cnt = 0;            while (n) {                int x = n % 10;                cnt += x * x;                n = n / 10;            }            if (cnt == 1) {                flag = true;                break;            }            n = cnt;        }        if (flag) return true;        return false;    }};

int digitSquareSum(int n) {    //评论区做法    int sum = 0, tmp;    while (n) {        tmp = n % 10;        sum += tmp * tmp;        n /= 10;    }    return sum;}bool isHappy(int n) {    int slow, fast;    slow = fast = n;    do {        slow = digitSquareSum(slow);        fast = digitSquareSum(fast);        fast = digitSquareSum(fast);    } while(slow != fast);    if (slow == 1) return 1;    else return 0;}