add Happy Number

billryan · billryan · commit a4dc714e54a2 · 2017-04-02T22:30:18.000+08:00
diff --git a/zh-hans/SUMMARY.md b/zh-hans/SUMMARY.md
@@ -99,6 +99,7 @@
    * [Update Bits](math_and_bit_manipulation/update_bits.md)
    * [Fast Power](math_and_bit_manipulation/fast_power.md)
    * [Hash Function](math_and_bit_manipulation/hash_function.md)
+   * [Happy Number](math_and_bit_manipulation/happy_number.md)
    * [Count 1 in Binary](math_and_bit_manipulation/count_1_in_binary.md)
    * [Fibonacci](math_and_bit_manipulation/fibonacci.md)
    * [A plus B Problem](math_and_bit_manipulation/a_plus_b_problem.md)
diff --git a/zh-hans/math_and_bit_manipulation/happy_number.md b/zh-hans/math_and_bit_manipulation/happy_number.md
@@ -0,0 +1,76 @@
+# Happy Number
+
+Tags: Hash Table, Math, Easy
+
+## Question
+
+- leetcode: [Happy Number](https://leetcode.com/problems/happy-number/)
+- lintcode: [Happy Number](http://www.lintcode.com/en/problem/happy-number/)
+
+### Problem Statement
+
+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
+
+**Credits:**  
+Special thanks to [@mithmatt](https://leetcode.com/discuss/user/mithmatt) and
+[@ts](https://leetcode.com/discuss/user/ts) for adding this problem and
+creating all test cases.
+
+## 题解
+
+根据指定运算规则判断输入整数是否为『happy number』，容易推断得知最终要么能求得1，要么为环形队列不断循环。
+第一种情况容易判断，第二种情况即判断得到的数是否为环形队列，也就是说是否重复出现，这种场景使用哈希表轻易解决。
+
+### Java
+
+```java
+public class Solution {
+    public boolean isHappy(int n) {
+        if (n < 0) return false;
+
+        Set<Integer> set = new HashSet<Integer>();
+        set.add(n);
+        while (n != 1) {
+            n = digitsSquareSum(n);
+            if (n == 1) {
+                return true;
+            } else if (set.contains(n)) {
+                return false;
+            } else {
+                set.add(n);
+            }
+        }
+
+        return true;
+    }
+
+    private int digitsSquareSum(int n) {
+        int sum = 0;
+        for (; n > 0; n /= 10) {
+            sum += (n % 10) * (n % 10);
+        }
+        return sum;
+    }
+}
+```
+
+### 源码分析
+
+辅助方法计算数字平方和。
+
+### 复杂度分析
+
+有限迭代次数一定终止，时间和空间复杂度均为 $$O(1)$$.
