My code:

public class Solution {
    public String countAndSay(int n) {
        if (n <= 0)
            return null;
        else if (n == 1)
            return "1";
        int count = 1;
        String result = "1";
        while (count < n) {
            int begin = 0;
            int i = begin + 1;
            String temp = "";
            while (i < result.length()) {
                if (result.charAt(i) == result.charAt(i - 1))
                    i++;
                else {
                    temp += Integer.toString(i - begin) + result.charAt(begin);
                    begin = i;
                    i++;
                }
            }
            temp += Integer.toString(i - begin) + result.charAt(begin);
            result = temp;
            count++;
        }
        return result;
    }
    
    public static void main(String[] args) {
        Solution test = new Solution();
        System.out.println(test.countAndSay(2));
        
    }
}

My test result:

Paste_Image.png

這次作業(yè)比較有趣〕鹾瘢可以科普下什么叫做 count and say number sequence.

https://en.wikipedia.org/wiki/Look-and-say_sequence

Paste_Image.png

也沒(méi)有什么技術(shù)含量件蚕。還是簡(jiǎn)單的掃描字符串。

**
總結(jié)： String, scan
**

Anyway, Good luck, Richardo!

My code:

public class Solution {
    public String countAndSay(int n) {
        if (n <= 0)
            return null;
        StringBuilder s = new StringBuilder(String.valueOf(1));
        for (int i = 1; i < n; i++) {
            s = helper(s);
        }
        return s.toString();
    }
    
    private StringBuilder helper(StringBuilder s) {
        if (s == null || s.length() == 0)
            return null;
        StringBuilder ret = new StringBuilder();
        int counter = 1;
        char pre = s.charAt(0);
        for (int i = 1; i < s.length(); i++) {
            char curr = s.charAt(i);
            if (curr != pre) {
                ret.append(String.valueOf(counter) + pre);
                counter = 1;
                pre = curr;
            }
            else {
                counter++;
            }
        }
        if (counter != 0) {
            ret.append(String.valueOf(counter) + pre);
        }
        return ret;
    }
}

一開(kāi)始拿String做的产禾，排名在 30%左右排作。換成，StringBuilder之后亚情，排名變成了70%
題目本身并沒(méi)有什么難度妄痪。
今天把zappos的OA做了。感覺(jué)不是很難楞件。但是也是提前看了題目之后才說(shuō)出這樣的話的∩郎現(xiàn)在看到string，palindrome罪针，parentheses黄伊。。。第一反應(yīng)就是愈腾，DP, GREEDY
其實(shí)完全不用這么緊張的。

palindrome 用DP
valid parentheses可以用dp虱黄，也可用stack

Anyway, Good luck, Richardo!

My code:

public class Solution {
    HashMap<Integer, String> map = new HashMap<Integer, String>();
    public String countAndSay(int n) {
        if (n <= 0) {
            return null;
        }
        String s = "1";
        for (int i = 1; i < n; i++) {
            s = transform(s);
        }
        return s;
    }
    
    private String transform(String s) {
        StringBuilder ret = new StringBuilder();
        int begin = 0;
        int end = 1;
        while (end <= s.length()) {
            if (end < s.length() && s.charAt(end) == s.charAt(end - 1)) {
                end++;
            }
            else {
                int diff = end - begin;
                ret.append(diff);
                ret.append(s.charAt(begin));
                begin = end;
                end++;
            }
        }
        return ret.toString();
    }
}

reference:
https://discuss.leetcode.com/topic/14543/straightforward-java-solution/2

這道題目我犯傻比了辜梳。
我的想法是：
比如輸入 Integer 11
我先寫(xiě)個(gè)函數(shù)泳叠，將Integer -> String
11 -> "Two 1"
再寫(xiě)個(gè)函數(shù)，將String -> Integer
"Tow 1" -> 21
然后我再拿 21 輸入第一個(gè)函數(shù)宗挥，以此類推，拿到第n個(gè)數(shù)

其實(shí)完全不用這么復(fù)雜契耿。因?yàn)槲覀冏詈笮枰敵龅模皇?"Two 1"
而是 "21"

所以透敌，直接把 "11" -> "21"
這樣踢械，一下就變簡(jiǎn)單題了。

Anyway, Good luck, Richardo! -- 09/20/2016