【leetcode】155. Min Stack

一、题目描述

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.

题目解读：写一个栈要有以下功能：push()，pop()，top()，getMin()能取得当前栈中的最小元素

思路：直接用一个栈就可以实现所有的push()，pop()，top()这三个功能，在设置一个栈用来存每加入一个元素时当前的最小元素。因此在每pop一个元素对于min栈也要pop一个元素。

c++代码（52ms，74.75%）

class MinStack {
public:
/** initialize your data structure here. */
stack<int> min;
stack<int> result;

void push(int x) {
result.push(x);
if(min.empty())
min.push(x);
else{
if(x<min.top())
min.push(x);
else{
min.push(min.top());
}//else
}//else
}//void

void pop() {
result.pop();
min.pop();
}

int top() {
return result.top();
}

int getMin() {
return min.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.getMin();
*/

其他代码：

class MinStack {
private:
stack<int> s1;
stack<int> s2;
public:
void push(int x) {
s1.push(x);
if (s2.empty() || x <= getMin()) s2.push(x);
}
void pop() {
if (s1.top() == getMin()) s2.pop();
s1.pop();
}
int top() {
return s1.top();
}
int getMin() {
return s2.top();
}
};