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;
}
};
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
原文：http://www.cnblogs.com/liyangguang1988/p/3984575.html