- Leetcode 20. Valid Parentheses
- Leetcode 22. Generate Parentheses
- Leetcode 32. Longest Valid Parentheses
- Leetcode 241. Different Ways to Add Parentheses
Parentheses Series
Leetcode 20. 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.
class Solution {
public:
bool isValid(string s) {
stack<char> st;
if (s.empty()) return true;
for (int i=0; i<s.size(); i++) {
char sign = s[i];
if (s[i]=='[' || s[i]=='(' || s[i]=='{')
st.push(sign);
else {
if (st.empty()) return false;
if (sign==')') {
if (st.top() != '(') return false;
else st.pop();
}
else if (sign==']') {
if (st.top() != '[') return false;
else st.pop();
}
else {
if (st.top() != '{') return false;
else st.pop();
}
}
}
return st.size()==0;
}
};
Leetcode 22. 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:
[
"((()))",
"(()())",
"(())()",
"()(())",
"()()()"
]
class Solution {
public:
vector<string> generateParenthesis(int n) {
vector<string> res;
if (n<=0) return res;
string str;
helper(res, str, 0, n);
return res;
}
void helper(vector<string> &res, string str, int right, int left) {
if (left==0 && right==0) {
res.push_back(str);
return;
}
if (right>0)
helper(res, str+")", right-1, left);
if (left>0)
helper(res, str+"(", right+1, left-1);
}
};
Leetcode 32. 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.
class Solution {
public:
int longestValidParentheses(string input) {
if (input.empty()) return 0;
stack<int> st;
int last = -1;
int res = 0;
for (int i=0; i<input.size(); i++) {
if (input[i]=='(') st.push(i);
else {
if (st.empty()) last = i;
else {
st.pop();
if (st.empty()) res = max(res, i-last);
else res = max(res, i-st.top());
}
}
}
return res;
}
};
Leetcode 241. Different Ways to Add Parentheses
Given a string of numbers and operators, return all possible results from computing all the different possible ways to group numbers and operators. The valid operators are +, - and *.
Example 1
Input: “2-1-1”.
((2-1)-1) = 0
(2-(1-1)) = 2
Output: [0, 2]
Example 2
Input: “23-45”
(2(3-(45))) = -34
((23)-(45)) = -14
((2(3-4))5) = -10
(2((3-4)5)) = -10
(((23)-4)5) = 10
Output: [-34, -14, -10, -10, 10]
class Solution {
public:
vector<int> diffWaysToCompute(string input) {
vector<int> res;
if (input.empty()) return res;
const int n = input.size();
for (int i=0; i<n; i++) {
char sign = input[i];
if (sign =='+' || sign=='-' || sign =='*') {
vector<int> temp1 = diffWaysToCompute(input.substr(0, i));
vector<int> temp2 = diffWaysToCompute(input.substr(i+1));
for (int num1:temp1) {
for (int num2:temp2) {
if (sign=='+') res.push_back(num1+num2);
else if (sign=='-') res.push_back(num1-num2);
else res.push_back(num1*num2);
}
}
}
}
if (res.empty())
res.push_back(stoi(input));
return res;
}
};
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