【第三轮】8.3-8.9 CareerCup 9.6

9.6 Implement an algorithm to print all valid (i.e., properly opened and closed) combinations of n-pairs of parentheses.


回复解法可以按照以下格式来

【解题思路】
【时间复杂度】
【空间复杂度】
【gist link】
---------------OPTional，如果觉得test case比较好，欢迎写出来分享----------------------
【test case】

【解题思路】
keep track of the number of left and right parentheses allowed.
both are zero: add to the result list
leftCap != 0: add a ( and recurse
rightCap > leftCap: add a ) and recurse
【时间复杂度】
O(2^2N) 2 recursive call for 2N layers
【空间复杂度】
O(2^2N)
【gist link】
https://gist.github.com/54f351172827af15c037

补充内容 (2014-8-10 22:00):
number of valid combinations: catalan number!

【解题思路】
if possible, add a left parenthese and then recurse
if possible, add a right parenthese and then recurse



【时间复杂度】
O(2^n)


【空间复杂度】
O(2^n)


【gist link】
https://gist.github.com/happyWinner/b5f342df67518297f704

【解题思路】
Recursion, with the check that number of available right parentheses should always be more than that of left parentheses

【时间复杂度】
O(2^N)

【空间复杂度】
O(2^N) for recursion stack only

【gist link】
https://gist.github.com/chrislukkk/b501cb96f9006170bc86

【解题思路】Recursion, if left has not been used up, insert (, then if left < right, insert )
【时间复杂度】
O(2^N)
【空间复杂度】
O(2^N)
【gist link】
https://gist.github.com/bitcpf/6bc2a883bd1c3df500ea