##题目
####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.
##解题思路
这道题实现一个最小栈的问题，这里通过压入每个元素的同时，在压入此时整个栈中的最小值。这样栈里的元素其实是压入元素与当前最小值间的交叉排列。
##算法代码
代码采用JAVA实现：
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
|
class {
LinkedList<Integer> stack=new LinkedList<Integer>();
public void push(int x) {
if(stack.isEmpty())
{
stack.push(x);
stack.push(x);
}else{
int min=stack.peek();
stack.push(x);
if(x>min)
{
stack.push(min);
}else{
stack.push(x);
}
}
}
public void pop() {
if(stack.isEmpty())
return;
else
{
stack.pop();
stack.pop();
}
}
public int top() {
if(stack.isEmpty())
return -1;
else{
int num=stack.pop();
int res=stack.peek();
stack.push(num);
return res;
}
}
public int getMin() {
if(stack.isEmpty())
return -1;
else
return stack.peek();
}
}
|
近期评论