add Count and Say

billryan · billryan · commit 2e23e89168e4 · 2015-09-17T21:51:48.000+08:00
diff --git a/zh-cn/SUMMARY.md b/zh-cn/SUMMARY.md
@@ -35,6 +35,7 @@
    * [Space Replacement](string/space_replacement.md)
    * [Wildcard Matching](string/wildcard_matching.md)
    * [Length of Last Word](string/length_of_last_word.md)
+   * [Count and Say](string/count_and_say.md)
 * [Integer Array](integer_array/README.md)
    * [Remove Element](integer_array/remove_element.md)
    * [Zero Sum Subarray](integer_array/zero_sum_subarray.md)
diff --git a/zh-cn/string/count_and_say.md b/zh-cn/string/count_and_say.md
@@ -0,0 +1,75 @@
+# Count and Say
+
+## Source
+
+- leetcode: [Count and Say | LeetCode OJ](https://leetcode.com/problems/count-and-say/)
+- lintcode: [(420) Count and Say](http://www.lintcode.com/en/problem/count-and-say/)
+
+```
+The count-and-say sequence is the sequence of integers beginning as follows:
+
+1, 11, 21, 1211, 111221, ...
+
+1 is read off as "one 1" or 11.
+
+11 is read off as "two 1s" or 21.
+
+21 is read off as "one 2, then one 1" or 1211.
+
+Given an integer n, generate the nth sequence.
+
+Example
+Given n = 5, return "111221".
+
+Note
+The sequence of integers will be represented as a string.
+```
+
+## 题解
+
+题目大意是找第 n 个数(字符串表示)，规则则是对于连续字符串，表示为重复次数+数本身。
+
+### Java
+
+```java
+public class Solution {
+    /**
+     * @param n the nth
+     * @return the nth sequence
+     */
+    public String countAndSay(int n) {
+        if (n <= 0) return null;
+
+        String s = "1";
+        for (int i = 1; i < n; i++) {
+            int count = 1;
+            StringBuilder sb = new StringBuilder();
+            int sLen = s.length();
+            for (int j = 0; j < sLen; j++) {
+                if (j < sLen - 1 && s.charAt(j) == s.charAt(j + 1)) {
+                    count++;
+                } else {
+                    sb.append(count + "" + s.charAt(j));
+                    // reset
+                    count = 1;
+                }
+            }
+            s = sb.toString();
+        }
+
+        return s;
+    }
+}
+```
+
+### 源码分析
+
+字符串是动态生成的，故使用 StringBuilder 更为合适。注意s 初始化为"1", 第一重 for循环中注意循环的次数为 n-1.
+
+### 复杂度分析
+
+略
+
+## Reference
+
+- [[leetcode]Count and Say - 喵星人与汪星人](http://huntfor.iteye.com/blog/2059877)
