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72. Generate Parentheses && Valid Parentheses
Generate 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。
思路：可用于卡特兰数一类题目。
void getParenthesis(vector<string> &vec, string s, int left, int right) { if(!right && !left) { vec.push_back(s); return; } if(left > 0) getParenthesis(vec, s+"(", left-1, right); if(right > 0 && left < right) getParenthesis(vec, s+")", left, right-1);}class Solution {public: vector<string> generateParenthesis(int n) { vector<string> vec; getParenthesis(vec, string(), n, n); return vec; }};
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.
思路： 栈。对 S 的每个字符检查栈尾，若成对，则出栈，否则，入栈。
class Solution {public: bool isValid(string s) { bool ans = true; char ch[6] = {‘(‘, ‘{‘, ‘[‘, ‘]‘, ‘}‘, ‘)‘}; int hash[256] = {0}; for(int i = 0; i < 6; ++i) hash[ch[i]] = i; string s2; for(size_t i = 0; i < s.size(); ++i) { if(s2 != "" && hash[s2.back()] + hash[s[i]] == 5) s2.pop_back(); else s2.push_back(s[i]); } if(s2 != "") ans = false; return ans; }};
72. Generate Parentheses && Valid Parentheses
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