leetcode 剑指 offer 面试题30. 包含min函数的栈

面试题30. 包含min函数的栈

本题与 leetcode 155 题相同：https://leetcode-cn.com/problems/min-stack/

不使用额外空间无法做到 O(1) 复杂度。
定义一个额外的栈 m 储存当前的最小值。

参考和详细解析：liweiwei1419 的题解

python 解法

class MinStack:

    def __init__(self):
        """
        initialize your data structure here.
        """
        self.s = []
        self.m = []


    def push(self, x: int) -> None:
        if not self.m or x <= self.m[-1]:
            self.m.append(x)
        self.s.append(x)


    def pop(self) -> None:
        x = self.s.pop()
        if x == self.m[-1]: 
            self.m.pop()
        return x


    def top(self) -> int:
        return self.s[-1]


    def min(self) -> int:
        return self.m[-1]



# Your MinStack object will be instantiated and called as such:
# obj = MinStack()
# obj.push(x)
# obj.pop()
# param_3 = obj.top()
# param_4 = obj.min()

c++ 解法

#include <stack>

class MinStack {
    stack<int> s, m;
public:
    /** initialize your data structure here. */
    MinStack() {

    }
    
    void push(int x) {
        s.push(x);
        if (m.empty() || x <= m.top()) {
            m.push(x);
        }
    }
    
    void pop() {
        if (m.top() == s.top()) {
            m.pop();
        }
        s.pop();
    }
    
    int top() {
        return s.top();
    }
    
    int min() {
        return m.top();
    }
};

/**
 * Your MinStack object will be instantiated and called as such:
 * MinStack* obj = new MinStack();
 * obj->push(x);
 * obj->pop();
 * int param_3 = obj->top();
 * int param_4 = obj->min();
 */