题目地址:https://leetcode.com/problems/happy-number/open in new window

Total Accepted: 36352 Total Submissions: 109782 Difficulty: Easy

题目描述

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

Ahappy 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:

Input: 19
Output: true
Explanation: 
1^2 + 9^2 = 82
8^2 + 2^2 = 68
6^2 + 8^2 = 100
1^2 + 0^2 + 0^2 = 1

题目大意

判断一个数字是不是开心的数字,所谓开心数字,就是把它的每一位数字求平方和之后构成新数字,然后继续这个操作,看最后能不能到1.

解题方法

递归

使用递归的方法。

我自己的算法,10以下的Happy Number 只有 1和7 ,如果一个数计算到只有个位数时,如果计算到十位以下,这个数是1或7,返回true,否则,返回false。

public static boolean isHappy(int n) {
	int ans = 0;
	if (n == 1 || n == 7) {
		return true;
	} else if (n > 1 && n < 10) {
		return false;
	} else {
		String numString = "" + n;
		char numChar[] = numString.toCharArray();
		for (char aNumChar : numChar) {
			ans += (aNumChar - '0') * (aNumChar - '0');
		}
	}
	return isHappy2(ans);
}

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方法一改进:

没必要10以下的数字啊,1到7之间的都是false。直接判断数到1和7之间 就false就好了。

7通过计算也回到1。

public static boolean isHappy(int n) {
	int ans = 0;
	if (n == 1) {
		return true;
	} else if (n > 1 && n < 7) {
		return false;
	} else {
		String numString = "" + n;
		char numChar[] = numString.toCharArray();
		for (char aNumChar : numChar) {
			ans += (aNumChar - '0') * (aNumChar - '0');
		}
	}
	return isHappy5(ans);
}

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迭代

同计算循环小数一样, 如果出现循环, 则无需继续计算,直接返回false即可.

每次计算时,把已经计算数放到一个集合里面,在计算过程中如果出现循环(集合里已经有这个数字),返回false。否则一直计算。

class Solution(object):
    def isHappy(self, n):
        """
        :type n: int
        :rtype: bool
        """
        visited = set()
        while n not in visited:
            visited.add(n)
            nx = 0
            while n != 0:
                nx += (n % 10) ** 2
                n //= 10
            if nx == 1:
                return True
            n = nx
        return False

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迭代的C++代码如下:

class Solution {
public:
    bool isHappy(int n) {
        unordered_set<int> visited;
        visited.insert(n);
        while (n != 1) {
            int pre = n;
            int next = 0;
            while (pre) {
                next += (pre % 10) * (pre % 10);
                pre /= 10;
            }
            n = next;
            if (visited.count(n))
                break;
            visited.insert(n);
        }
        return n == 1;
    }
};

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