题解 | #包含min函数的栈#
包含min函数的栈
http://www.nowcoder.com/practice/4c776177d2c04c2494f2555c9fcc1e49
方案1：借助双栈来解决
import java.util.*;
public class Solution {
Stack<Integer> normal=new Stack();
Stack<Integer> minVal=new Stack();
public void push(int value){
if(minVal.isEmpty()||value<minVal.peek()){
minVal.push(value);
}else{
minVal.push(minVal.peek());
}
normal.push(value);
}
public void pop(){
if(!normal.isEmpty()&&!minVal.isEmpty()){
normal.pop();
minVal.pop();
}
}
public int top(){
int res=-1;
if(!normal.isEmpty()){
res=normal.peek();
}
return res;
}
public int min(){
int res=-1;
if(!minVal.isEmpty()){
res=minVal.peek();
}
return res;
}
}
方案2：借助栈和优先级队列来解决
import java.util.*;
public class Solution {
Stack<Integer> s=new Stack();
Queue<Integer> q=new PriorityQueue();
public void push(int node) {
s.push(node);
q.offer(node);
}
public void pop() {
if(!s.isEmpty()){
q.remove(s.pop());
}
}
public int top() {
if(!s.isEmpty()){
return s.peek();
}
return -1;
}
public int min() {
return q.peek();
}
}
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