【第四轮】3.30-4.5 Career Cup 1.5

【解题思路】First thing is to see whether the compressed string will be larger than the original string. If so, return the original string. If not, traverse the string and count the unique characters and their frequency. concatenate them into anew string.
【时间复杂度】O(n+k^2) n is the length og the original string, k is the number of unique characters in the original string
【空间复杂度】O(k)
【gist link] https://github.com/StephenWeiXu/ ... blob/master/1_5.cpp
【test case】(optional，如果觉得比较好，欢迎贴出来分享）
https://github.com/StephenWeiXu/ ... blob/master/1_5.cpp

【解题思路】
* When we want to add something to a string constantly and without knowing the total length,
* StringBuilder is a good choice.
* We check whether the adjacent elements are same. if so, we increase count number; if not, we output the count number,
* new character and reset the count number to 1.
* Please be aware that at the end, we cannot detect the different adjacent elements, but we still need to add the count number.

【时间复杂度】
O(n)
【空间复杂度】
O(n)
【gist link]
https://gist.github.com/leonw007/ef69b87b8ec03c1a7d08
【test case】included in the above link

【解题思路】
扫一遍如果相同count++,如果不同，count重置为1，同时把之前的count append到之前的字符后。

【时间复杂度】
O(n)
【空间复杂度】
O(n)
【gist link]
https://gist.github.com/songpu2015617/7b1e335600a5e5986518

[solution]
scan the string, use a prev value and a counter to keep track of the repeat of a char, and then update the compressed string accordingly.
[time]
O(n)
[space]
O(n)
[gist]
https://gist.github.com/kelly-us/b1777dc1dc582cf253ad

【解题思路】
Tips:
* Use two pointer to make the begin of repeat String snippet and current char, and
* a counter to mark the num of the repeat String snippet
【时间复杂度】O(N)
【空间复杂度】O(N)
【gist link】https://github.com/drinkbeer/Cod ... /src/CC150_1_5.java
【test case】