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
- 1^2 + 9^2 = 82
- 8^2 + 2^2 = 68
- 6^2 + 8^2 = 100
- 1^2 + 0^2 + 0^2 = 1
Credits:
Special thanks to @mithmatt and @ts for adding this problem and creating all test cases.
寫一個算法判斷一個數是不是快樂數。快樂數:一個正整數,如果對其各個位上的數字分別平方求和得到一個新的數,再進行同樣的操作,如果結果變成了1,則說明是快樂數。
解法:循環求平方和,即求出當前數組的平方和后,再以此平方和作為新的數繼續求平方和,而循環終止條件是:得到的平方和為1,或得到的平方和在之前的循環中出現過,那以后會一直循環,不可能達到1。判斷平方和是否為1很簡單,每次檢查就好了;而判斷平方和是否出現過,則只需要維持一個Set,每次循環檢查當前平方和是否在Set中,在則終止循環,不在則將此平方和放到Set中。
Java:
public class Solution {
public boolean isHappy(int n) {
Set<Integer> got = new HashSet<>();
while (n != 1 && !got.contains(n)) {
got.add(n);
int sum = 0;
while (n != 0) {
sum += Math.pow(n % 10, 2);
n /= 10;
}
n = sum;
}
return n == 1;
}
}
Python:
class Solution:
# @param {integer} n
# @return {boolean}
def isHappy(self, n):
lookup = {}
while n != 1 and n not in lookup:
lookup[n] = True
n = self.nextNumber(n)
return n == 1
def nextNumber(self, n):
new = 0
for char in str(n):
new += int(char)**2
return new
Python:
class Solution(object):
def isHappy(self, n):
"""
:type n: int
:rtype: bool
"""
got = set()
while n != 1 and n not in got:
got.add(n)
sum = 0
while n:
sum += (n % 10)**2
n //= 10
n = sum
return n == 1
C++:
class Solution {
public:
bool isHappy(int n) {
set<int> got;
while (n != 1 && got.find(n) == got.end()) {
got.insert(n);
int sum = 0;
while (n) {
sum += pow(n % 10, 2);
n /= 10;
}
n = sum;
}
return n == 1;
}
};
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