Min Stack 解答

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Question

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.

Solution

Original thinking is to use two stacks, one to store input elements, and the other is to store current min value.

However, there is an improvement that only use one stack.

This stack is to store diff between input value and current min value.

if current value < min value, we set min as current imput value.

Therefore, when we pop or peek each element in stack, we know:

If it‘s greater than 0, then it must be greater than current min value.

If it‘s smaller than 0, then it must equal to current min value.

 1 class MinStack {
 2     Stack<Long> diff;
 3     private long min;
 4     
 5     public MinStack() {
 6         min = Integer.MAX_VALUE;
 7         diff = new Stack<Long>();
 8     }
 9     public void push(int x) {
10         diff.push((long)x - min);
11         min = x < min? x : min;
12     }
13     public void pop() {
14         if (diff.size() < 1)
15             return;
16         long tmp = diff.pop();
17         if (tmp < 0)
18             min -= tmp;
19     }
20     public int top() {
21         long tmp = diff.peek();
22         if (tmp < 0)
23             tmp = min;
24         else
25             tmp += min;
26         return (int)tmp;
27     }
28     public int getMin() {
29         return (int)min;
30     }
31 }

Min Stack 解答

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原文地址：http://www.cnblogs.com/ireneyanglan/p/4860010.html