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leetcode 之 Longest Valid Parentheses

leetcode中和括号匹配相关的问题共有三个,分别是:

Valid Parentheses

 

Given a string containing just the characters ‘(‘‘)‘‘{‘‘}‘‘[‘ and ‘]‘, determine if the input string is valid.

The brackets must close in the correct order, "()" and "()[]{}" are all valid but "(]" and "([)]" are not.

该提比较简单,正常情况下直接用堆栈就可以了,但有一次面试要求必须要用递归写,其实也很简单,具体参考这里


Longest Valid Parentheses

 

Given a string containing just the characters ‘(‘ and ‘)‘, find the length of the longest valid (well-formed) parentheses substring.

For "(()", the longest valid parentheses substring is "()", which has length = 2.

Another example is ")()())", where the longest valid parentheses substring is "()()", which has length = 4.

该题目使用动态规划来计算,dp[i]表示到第i个位置的最大长度,由于匹配的括号必须是连续的,所以,如果有j < i 且j和i匹配,则dp[i] = (i-j+)+dp[j]。

从转移方程来看,好像是二维DP,但是可以使用堆栈来转化为一维的,简单来说,就是遇到左括号就进栈,遇到右括号就出栈,而出栈的位置就是上

面的j,所以不需要进行二维扫描就可定位到j。

class Solution {
public:
    int longestValidParentheses(string s) {
    	int length = s.size(),i,maxLength = 0;
    	vector<int> dp(length,0);
    	stack<int> stk; // 左括号的下标
    	for(i = 0; i < length;++i)
    	{
    		if(s[i] == '(')stk.push(i);
    		else
    		{
    			if(!stk.empty())
    			{
    				int start = stk.top();
    				stk.pop();
    				dp[i] = i - start + 1;
    				if(start > 0)dp[i] += dp[start-1];
    				if(dp[i] > maxLength)maxLength = dp[i];
    			}
    		}
    	}
    	return maxLength;
    }
};


Generate Parentheses

  

Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

For example, given n = 3, a solution set is:

"((()))", "(()())", "(())()", "()(())", "()()()"

该问题是著名的卡特兰数,具体参考该博客