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:
Input: 19
Output: true
Explanation:
12 + 92 = 82
82 + 22 = 68
62 + 82 = 100
12 + 02 + 02 = 1
思路
（1）剛開始想的是動態規劃，當結果是1,10,100,1000…的情況下都可以，然後1+9=10，36+64=100，這樣就想到利用記憶表來自底向上的動態規劃。類似題：LeetCode-BFS&DP-279-M：完全平方數（Perfect Squares）
（2）關鍵是沒有讀懂repeat the process until the number equals 1 (where it will stay), or it loops endlessly in a cycle which does not include 1，重複這個過程直到這個數變爲 1，也可能是無限循環但始終變不到 1。說明結果只有兩種，要麼出現1，要麼無限循環，不會有無限不循環的情況。
（3）解法1利用哈希表來記錄處理，遇到重複數字就返回false，否則必定可以遇到1，跳出循環。new HashSet<>()
（4）解法2快慢指針。
解法1-哈希
class Solution {
public boolean isHappy(int n) {
if(n == 1){
return true;
}
HashSet<Integer> set = new HashSet<>();
while(n != 1){
int tmp = 0;
while(n > 9){
int a = n%10;
tmp += a*a;
n /=10;
}
tmp += n*n;
n = tmp;
if(set.contains(n)){
return false;
}else{
set.add(n);
}
}
return true;
}
}
解法2-快慢指針
class Solution {
public boolean isHappy(int n) {
int fast=n;
int slow=n;
do{
slow=squareSum(slow);
fast=squareSum(fast);
fast=squareSum(fast);
}while(slow!=fast);
if(fast==1)
return true;
else return false;
}
private int squareSum(int m){
int squaresum=0;
while(m!=0){
squaresum+=(m%10)*(m%10);
m/=10;
}
return squaresum;
}
}