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Commit 68f8b47

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Merge pull request doocs#147 from Wushiyii/mysolu-leetcode
Add solution of 202.Happy Number(java)
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‎solution/0202.Happy Number/README.md‎

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## 快乐数
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### 题目描述
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编写一个算法来判断一个数是不是"快乐数"。
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一个"快乐数"定义为:
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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。如果可以变为 1,那么这个数就是快乐数。
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示例:
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```
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输入: 19
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输出: true
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解释:
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12 + 92 = 82
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82 + 22 = 68
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62 + 82 = 100
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12 + 02 + 02 = 1
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```
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### 解法
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在进行验算的过程中,发现一个规律,只要过程中得到任意一个结果和为4,那么就一定会按 `4 → 16 → 37 → 58 → 89 → 145 → 42 → 20 → 4`
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进行循环,这样的数就不为快乐数;此外,结果和与若是与输入n或者上一轮结果和n相同,那也不为快乐数.
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```java
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class Solution {
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public boolean isHappy(int n) {
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if (n <= 0) return false;
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int sum = 0;
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while (sum != n) {
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while (n > 0) {
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sum += Math.pow(n % 10 ,2);
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n /= 10;
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}
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if (sum == 1) {
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return true;
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} else if (sum == 4) {
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return false;
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} else {
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n = sum;
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sum = 0;
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}
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}
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return false;
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}
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}
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// 递归
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public boolean isHappy(int n) {
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if (n <= 0) return false;
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int sum = 0;
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while (n > 0) {
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sum += Math.pow(n % 10 ,2);
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n /= 10;
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}
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if (sum == 1) {
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return true;
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} else if (sum == 4) {
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return false;
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} else {
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return isHappy(sum);
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}
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}
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```
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class Solution {
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public boolean isHappy(int n) {
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if (n <= 0) return false;
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int sum = 0;
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while (sum != n) {
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while (n > 0) {
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sum += Math.pow(n % 10, 2);
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n /= 10;
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}
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if (sum == 1) {
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return true;
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} else if (sum == 4) {
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return false;
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} else {
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n = sum;
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sum = 0;
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}
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}
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return false;
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}
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// 递归
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public boolean isHappy2(int n) {
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if (n <= 0) return false;
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int sum = 0;
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while (n > 0) {
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sum += Math.pow(n % 10, 2);
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n /= 10;
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}
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if (sum == 1) {
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return true;
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} else if (sum == 4) {
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return false;
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} else {
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return isHappy2(sum);
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}
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}
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}

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