标签：repeat several none als step java hit set algorithm
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
很自然的解法是：
class Solution(object):
def isHappy(self, n):
"""
:type n: int
:rtype: bool
"""
def sum2(num):
ans = 0
while num != 0:
ans += (num % 10)**2
num = num/10
return ans
m = set()
while n not in m:
if n == 1:
return True
m.add(n)
n = sum2(n)
return False
还有可以使用环路链表的判断方法，一个快慢指针一起赛跑，直到追上：
class Solution(object):
def isHappy(self, n):
"""
:type n: int
:rtype: bool
"""
def sum2(num):
ans = 0
while num != 0:
ans += (num % 10)**2
num = num/10
return ans
slow = n
fast = sum2(sum2(n))
while slow != fast:
slow = sum2(slow)
fast = sum2(sum2(fast))
return slow == 1
java代码其实非常优雅：
int digitSquareSum(int n) {
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;
}
数学解法：
class Solution {
public:
int loop[8] = {4,16,37,58,89,145,42,20};
bool inLoop(int n){
for(auto x: loop){
if(x == n) return true;
}
return false;
}
bool isHappy(int n) {
if(n == 1) return true;
if(inLoop(n)) return false;
int next = 0;
while(n){
next += (n%10)*(n%10);
n /= 10;
}
return isHappy(next);
}
};
proof:
1.loop number is less than 162.
Assume f(x) is the sum of the squares of x’s digits. let’s say 0 < x <= 9,999,999,999 which is larger than the range of an int. f(x) <= 9^2 * 10 = 810. So no mater how big x is, after one step of f(x), it will be less than 810.The most large f(x) value (x < 810) is f(799) = 211. And do this several times: f(211) < f(199) = 163. f(163) < f(99) = 162. So no mater which x you choose after several times of f(x),it finally fall in the range of [1,162] and can never get out.
2.I check every unhappy number in range of [1,162] , they all fall in loop {4,16,37,58,89,145,42,20} ,which means every unhappy number fall in this loop.
其实通过枚举就应该知道上述<=810范围内的数字都会落在特定范围，通过不断缩小范围来找规律。=》
Using fact all numbers in [2, 6] are not happy (and all not happy numbers end on a cycle that hits this interval):
bool isHappy(int n) {
while(n>6){
int next = 0;
while(n){next+=(n%10)*(n%10); n/=10;}
n = next;
}
return n==1;
}
class Solution(object):
def isHappy(self, n):
"""
:type n: int
:rtype: bool
"""
def sum2(num):
ans = 0
while num != 0:
ans += (num % 10)**2
num = num/10
return ans
while n > 6:
n = sum2(n)
return n == 1
标签：repeat several none als step java hit set algorithm
原文地址：https://www.cnblogs.com/bonelee/p/8799337.html
