LeetCode-Min Stack
Min Stack
##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
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49class MinStack {
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();
}
}