202. 快乐数

English Version

题目描述

编写一个算法来判断一个数 n 是不是快乐数。

「快乐数」 定义为：

• 对于一个正整数，每一次将该数替换为它每个位置上的数字的平方和。
• 然后重复这个过程直到这个数变为 1，也可能是 无限循环 但始终变不到 1。
• 如果这个过程 结果为 1，那么这个数就是快乐数。

如果 n 是 快乐数 就返回 true ；不是，则返回 false 。

 

示例 1：

输入：n = 19
输出：true
解释：
12 + 92 = 82
82 + 22 = 68
62 + 82 = 100
12 + 02 + 02 = 1

示例 2：

输入：n = 2
输出：false

 

提示：

• 1 <= n <= 231 - 1

解法

方法一：哈希表 + 模拟

将每次转换后的数字存入哈希表，如果出现重复数字，说明进入了循环，不是快乐数。否则，如果转换后的数字为 $1$，说明是快乐数。

时间复杂度 $O(\log n)$，空间复杂度 $O(\log n)$。

class Solution:
    def isHappy(self, n: int) -> bool:
        vis = set()
        while n != 1 and n not in vis:
            vis.add(n)
            x = 0
            while n:
                n, v = divmod(n, 10)
                x += v * v
            n = x
        return n == 1

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

class Solution {
public:
    bool isHappy(int n) {
        unordered_set<int> vis;
        while (n != 1 && !vis.count(n)) {
            vis.insert(n);
            int x = 0;
            for (; n; n /= 10) {
                x += (n % 10) * (n % 10);
            }
            n = x;
        }
        return n == 1;
    }
};

func isHappy(n int) bool {
	vis := map[int]bool{}
	for n != 1 && !vis[n] {
		vis[n] = true
		x := 0
		for ; n > 0; n /= 10 {
			x += (n % 10) * (n % 10)
		}
		n = x
	}
	return n == 1
}

function isHappy(n: number): boolean {
    const getNext = (n: number) => {
        let res = 0;
        while (n !== 0) {
            res += (n % 10) ** 2;
            n = Math.floor(n / 10);
        }
        return res;
    };
    const set = new Set();
    while (n !== 1) {
        const next = getNext(n);
        if (set.has(next)) {
            return false;
        }
        set.add(next);
        n = next;
    }
    return true;
}

use std::collections::HashSet;
impl Solution {
    fn get_next(mut n: i32) -> i32 {
        let mut res = 0;
        while n != 0 {
            res += (n % 10).pow(2);
            n /= 10;
        }
        res
    }

    pub fn is_happy(mut n: i32) -> bool {
        let mut set = HashSet::new();
        while n != 1 {
            let next = Self::get_next(n);
            if set.contains(&next) {
                return false;
            }
            set.insert(next);
            n = next;
        }
        true
    }
}

int getNext(int n) {
    int res = 0;
    while (n) {
        res += (n % 10) * (n % 10);
        n /= 10;
    }
    return res;
}

bool isHappy(int n) {
    int slow = n;
    int fast = getNext(n);
    while (slow != fast) {
        slow = getNext(slow);
        fast = getNext(getNext(fast));
    }
    return fast == 1;
}

方法二：快慢指针

与判断链表是否存在环原理一致。如果 $n$ 是快乐数，那么快指针最终会与慢指针相遇，且相遇时的数字为 $1$；否则，快指针最终会与慢指针相遇，且相遇时的数字不为 $1$。

因此，最后判断快慢指针相遇时的数字是否为 $1$ 即可。

时间复杂度 $O(\log n)$，空间复杂度 $O(1)$。

class Solution:
    def isHappy(self, n: int) -> bool:
        def next(x):
            y = 0
            while x:
                x, v = divmod(x, 10)
                y += v * v
            return y

        slow, fast = n, next(n)
        while slow != fast:
            slow, fast = next(slow), next(next(fast))
        return slow == 1

class Solution {
    public boolean isHappy(int n) {
        int slow = n, fast = next(n);
        while (slow != fast) {
            slow = next(slow);
            fast = next(next(fast));
        }
        return slow == 1;
    }

    private int next(int x) {
        int y = 0;
        for (; x > 0; x /= 10) {
            y += (x % 10) * (x % 10);
        }
        return y;
    }
}

class Solution {
public:
    bool isHappy(int n) {
        auto next = [](int x) {
            int y = 0;
            for (; x; x /= 10) {
                y += pow(x % 10, 2);
            }
            return y;
        };
        int slow = n, fast = next(n);
        while (slow != fast) {
            slow = next(slow);
            fast = next(next(fast));
        }
        return slow == 1;
    }
};

func isHappy(n int) bool {
	next := func(x int) (y int) {
		for ; x > 0; x /= 10 {
			y += (x % 10) * (x % 10)
		}
		return
	}
	slow, fast := n, next(n)
	for slow != fast {
		slow = next(slow)
		fast = next(next(fast))
	}
	return slow == 1
}

function isHappy(n: number): boolean {
    const getNext = (n: number) => {
        let res = 0;
        while (n !== 0) {
            res += (n % 10) ** 2;
            n = Math.floor(n / 10);
        }
        return res;
    };

    let slow = n;
    let fast = getNext(n);
    while (slow !== fast) {
        slow = getNext(slow);
        fast = getNext(getNext(fast));
    }
    return fast === 1;
}

impl Solution {
    pub fn is_happy(n: i32) -> bool {
        let get_next = |mut n: i32| {
            let mut res = 0;
            while n != 0 {
                res += (n % 10).pow(2);
                n /= 10;
            }
            res
        };
        let mut slow = n;
        let mut fast = get_next(n);
        while slow != fast {
            slow = get_next(slow);
            fast = get_next(get_next(fast));
        }
        slow == 1
    }
}