[Leetcode] 155. Min Stack 解题报告

题目：

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

• push(x) -- Push element x onto stack.
• pop() -- Removes the element on top of the stack.
• top() -- Get the top element.
• getMin() -- Retrieve the minimum element in the stack.

Example:

MinStack minStack = new MinStack();minStack.push(-2);minStack.push(0);minStack.push(-3);minStack.getMin();   --> Returns -3.minStack.pop();minStack.top();      --> Returns 0.minStack.getMin();   --> Returns -2.

思路：

我自己的实现思路是：在栈中存贮一个pair<int, int>，其中第一个int是原有栈中的数值，第二个int是从栈底到当前位置的所有数值中的最小值。这种实现的空间复杂度是O(n)，getMin()时间复杂度是O(1)。如果还是用原有的栈，只是在getMin()方法的实现中采用线性遍历，那么空间复杂度是O(1)，时间复杂度则上升为O(n)。不知道有没有更好地方法？

代码：

class MinStack {public:    /** initialize your data structure here. */    MinStack() {    }        void push(int x) {        int min_value = st.empty() ? x : min(x, st.top().second);        st.push(make_pair(x, min_value));    }        void pop() {        st.pop();    }        int top() {        return st.top().first;    }        int getMin() {        return st.top().second;    }private:    stack<pair<int, int>> st;   // in the pair, the first is the value, and the second is the min value up to now};/** * 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.getMin(); */