leetcode 155

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         else
17         {
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  */