#leetcode#202. Happy Number
2017-02-20 15:31
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https://leetcode.com/problems/happy-number/
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
--------------------------------------------------------------------------------
如何判断happy number逻辑题目已经交代的很清楚了,就是对数字的每一位digit的平方求和, 看这个和是否能为1. 注意有可能出现无限循环,所以用一个hashset来看当前平方的和有没有出现过,如果有的话就会无限循环
public class Solution {
public boolean isHappy(int n) {
if(n <= 0){
return false;
}
Set<Integer> set = new HashSet<>(); // to detect endless loop
while(true){
int sum = 0;//put it in the loop
while(n > 0){
int curDigit = n % 10;
n /= 10;
sum += curDigit * curDigit;
}
if(sum == 1)
return true;
else if(!set.add(sum)){
return false;
}else{
n = sum;
}
}
}
}
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
--------------------------------------------------------------------------------
如何判断happy number逻辑题目已经交代的很清楚了,就是对数字的每一位digit的平方求和, 看这个和是否能为1. 注意有可能出现无限循环,所以用一个hashset来看当前平方的和有没有出现过,如果有的话就会无限循环
public class Solution {
public boolean isHappy(int n) {
if(n <= 0){
return false;
}
Set<Integer> set = new HashSet<>(); // to detect endless loop
while(true){
int sum = 0;//put it in the loop
while(n > 0){
int curDigit = n % 10;
n /= 10;
sum += curDigit * curDigit;
}
if(sum == 1)
return true;
else if(!set.add(sum)){
return false;
}else{
n = sum;
}
}
}
}
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