1，Valid Parentheses

 1 bool isVaild1(string& s) {  // 直接列举，不易扩展
 2     stack<char> stk;
 3     for (int i = 0; i < s.length(); ++i) {
 4         if (stk.empty())
 5             stk.push(s[i]);
 6         else {
 7             char top = stk.top();
 8             if (top == '(' && s[i] == ')' || top == '{' && s[i] == '}' || top == '[' && s[i] == ']')
 9                 stk.pop();
10         }
11     }
12     if (stk.empty())
13         return true;
14     else
15         return false;
16 }
17 
18 bool isValid2(string& s) {
19     string left = "([{";
20     string right = ")]}";
21     stack<char> stk;
22 
23     for (auto c = s.begin(); c != s.end(); ++c) {
24         if (left.find(*c) != string::npos)
25             stk.push(*c);
26         else {
27             if (stk.empty() || stk.top() != left[right.find(*c)])
28                 return false;
29             else
30                 stk.pop();
31         }
32     }
33     return stk.empty();
34 }

isVaild

2，Longest Valid Parentheses

 1 int longestValidParentheses(const string& s) {
 2     int len = 0;
 3     stack<char> stk;
 4 
 5     for (size_t i = 0; i < s.size(); ++i) {
 6         if (stk.empty())
 7             stk.push(s[i]);
 8         else {
 9             if (stk.top() == '(' && s[i] == ')') {
10                 stk.pop();
11                 ++len;
12             }
13             else
14                 stk.push(s[i]);
15         }
16     }
17     return 2 * len;
18 }

longestValidParentheses

3，Largest Rectangle in Histogram

 1 int longestRectangleArea1(vector<int>& heights) {  // 暴力求解
 2     if (heights.size() == 0) return 0;
 3     int result = 0;
 4     for (int i = 0; i < heights.size(); ++i) {
 5         int minHeight = heights[i];
 6         if (i == heights.size() - 1 || heights[i]>heights[i + 1]) {  // 简单优化
 7             for (int j = i; j >= 0; --j) {
 8                 minHeight = min(minHeight,heights[j]);
 9                 result = max((i - j + 1)*minHeight, result);
10             }
11         }
12     }
13     return result;
14 }
15 
16 int longestRectangleArea2(vector<int>& heights) {  // 用 stack实现，未看懂
17     stack<int> s;
18     heights.push_back(0);
19     int result = 0;
20     for (int i = 0; i < heights.size();) {
21         if (s.empty() || heights[i] > heights[s.top()])
22             s.push(i++);
23         else {
24             int temp = s.top();
25             s.pop();
26             result = max(result, heights[temp] * (s.empty() ? i : i - s.top() - 1));
27         }
28     }
29     return result;
30 
31 }

longestRectangleArea

4，Evaluate Reverse Polish Notation

 1 int evalRPN(vector<string>& tokens) {
 2     stack<int> stk;
 3     string options = "+-*/";
 4     for (int i = 0; i < tokens.size(); ++i) {
 5         if (options.find(tokens[i]) == string::npos)  // 不是运算符
 6             stk.push(atoi(tokens[i].c_str()));
 7         else {
 8             int num1 = stk.top();
 9             stk.pop();
10             int num2 = stk.top();
11             stk.pop();
12 
13             if (tokens[i] == "+") stk.push(num1 + num2);
14             else if (tokens[i] == "-") stk.push(num1 - num2);
15             else if (tokens[i] == "*") stk.push(num1 * num2);
16             else stk.push(num1 / num2);
17         }
18     }
19     return stk.top();
20 }

evalRPN

 

 

 

以上题目来源于：https://github.com/soulmachine/leetcode(leetcode-cpp.pdf)