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leetcode 155

2024-08-16 06:37:05   210人阅读

155. Min Stack

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.

设计支持push、pop、top和在常量时间内检索最小元素的栈。 
push(x) —— 推送元素X进栈 
pop() —— 移除栈顶元素
top() —— 得到栈顶元素
getMin() —— 检索栈的最小元素
利用两个vector来实现，一个实现正常的栈，一个在压栈的时候存储最小元素，在检索栈的最小元素的时候直接取最小栈的栈顶元素即可。
代码如下：
 1 class MinStack { 2 public: 3     /** initialize your data structure here. */ 4     vector<int> allstack; 5     vector<int> minstack; 6     MinStack() { 7          8     } 9     10     void push(int x) {11         if(allstack.empty())12         {13             allstack.push_back(x);14             minstack.push_back(x);15         }16         else17         {18             allstack.push_back(x);19             if(x <= minstack[minstack.size()-1])20             {21                 minstack.push_back(x);22             }23         }24     }25     26     void pop() {27         if(allstack[allstack.size()-1] == minstack[minstack.size()-1])28         {29             minstack.pop_back();30         }31         allstack.pop_back();32     }33     34     int top() {35         return allstack[allstack.size()-1];36     }37     38     int getMin() {39         return minstack[minstack.size()-1];40     }41 };42 43 /**44  * Your MinStack object will be instantiated and called as such:45  * MinStack obj = new MinStack();46  * obj.push(x);47  * obj.pop();48  * int param_3 = obj.top();49  * int param_4 = obj.getMin();50  */
 


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