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155.min-stack.cpp
100 lines (91 loc) · 1.95 KB
/
155.min-stack.cpp
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/*
* @lc app=leetcode id=155 lang=cpp
*
* [155] Min Stack
*
* https://leetcode.com/problems/min-stack/description/
*
* algorithms
* Easy (35.66%)
* Total Accepted: 273.7K
* Total Submissions: 763.7K
* Testcase Example: '["MinStack","push","push","push","getMin","pop","top","getMin"]\n[[],[-2],[0],[-3],[],[],[],[]]'
*
*
* 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.
*
*
*/
// 解决思路
// 首先栈本身的性质就有push,pop,top三个功能
// 要解决线性时间取最小值的方案,想到的就是用另外一个栈来保存当前最小的值
class MinStack {
public:
/** initialize your data structure here. */
MinStack() {
}
void push(int x) {
nums.push(x);
if ( mins.empty() ) {
mins.push(x);
}else
{
int min = getMin();
if ( x < min) {
mins.push(x);
}
else
{
mins.push(min);
}
}
}
void pop() {
nums.pop();
mins.pop();
}
int top() {
return nums.top();
}
int getMin() {
return mins.top();
}
private:
stack<int> nums;
stack<int> mins;
};
/**
* 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();
*/