leetcode 栈和队列类型题

1，Valid Parentheses

bool isVaild1(string& s) {  // 直接列举，不易扩展
    stack<char> stk;
    for (int i = 0; i < s.length(); ++i) {
        if (stk.empty())
            stk.push(s[i]);
        else {
            char top = stk.top();
            if (top == '(' && s[i] == ')' || top == '{' && s[i] == '}' || top == '[' && s[i] == ']')
                stk.pop();
        }
    }
    if (stk.empty())
        return true;
    else
        return false;
}

bool isValid2(string& s) {
    string left = "([{";
    string right = ")]}";
    stack<char> stk;

    for (auto c = s.begin(); c != s.end(); ++c) {
        if (left.find(*c) != string::npos)
            stk.push(*c);
        else {
            if (stk.empty() || stk.top() != left[right.find(*c)])
                return false;
            else
                stk.pop();
        }
    }
    return stk.empty();
}

2，Longest Valid Parentheses

int longestValidParentheses(const string& s) {
    int len = 0;
    stack<char> stk;

    for (size_t i = 0; i < s.size(); ++i) {
        if (stk.empty())
            stk.push(s[i]);
        else {
            if (stk.top() == '(' && s[i] == ')') {
                stk.pop();
                ++len;
            }
            else
                stk.push(s[i]);
        }
    }
    return 2 * len;
}

3，Largest Rectangle in Histogram

int longestRectangleArea1(vector<int>& heights) {  // 暴力求解
    if (heights.size() == 0) return 0;
    int result = 0;
    for (int i = 0; i < heights.size(); ++i) {
        int minHeight = heights[i];
        if (i == heights.size() - 1 || heights[i]>heights[i + 1]) {  // 简单优化
            for (int j = i; j >= 0; --j) {
                minHeight = min(minHeight,heights[j]);
                result = max((i - j + 1)*minHeight, result);
            }
        }
    }
    return result;
}

int longestRectangleArea2(vector<int>& heights) {  // 用 stack实现，未看懂
    stack<int> s;
    heights.push_back(0);
    int result = 0;
    for (int i = 0; i < heights.size();) {
        if (s.empty() || heights[i] > heights[s.top()])
            s.push(i++);
        else {
            int temp = s.top();
            s.pop();
            result = max(result, heights[temp] * (s.empty() ? i : i - s.top() - 1));
        }
    }
    return result;

}

4，Evaluate Reverse Polish Notation

int evalRPN(vector<string>& tokens) {
    stack<int> stk;
    string options = "+-*/";
    for (int i = 0; i < tokens.size(); ++i) {
        if (options.find(tokens[i]) == string::npos)  // 不是运算符
            stk.push(atoi(tokens[i].c_str()));
        else {
            int num1 = stk.top();
            stk.pop();
            int num2 = stk.top();
            stk.pop();

            if (tokens[i] == "+") stk.push(num1 + num2);
            else if (tokens[i] == "-") stk.push(num1 - num2);
            else if (tokens[i] == "*") stk.push(num1 * num2);
            else stk.push(num1 / num2);
        }
    }
    return stk.top();
}

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