題目描述
定義棧的數(shù)據(jù)結(jié)構(gòu)淮蜈，請在該類型中實現(xiàn)一個能夠得到棧最小元素的min函數(shù)。
import java.util.Stack;
public class Solution {
//數(shù)據(jù)棧
Stack<Integer> s1 = new Stack<>();
//輔助棧
Stack<Integer> s2 = new Stack<>();
public void push(int node) {
s1.push(node);
if(s2.isEmpty()){
s2.push(node);
}else if(node<s2.peek()){
s2.push(node);
}else{
s2.push(s2.peek());
}
}
public void pop() {
s1.pop();
s2.pop();
}
public int top() {
return s1.peek();
}
public int min() {
return s2.peek();
}
}
2017.6.1 第二次做侄泽，思路略有變化
import java.util.Stack;
public class Solution {
Stack<Integer> stack1 = new Stack<>();
Stack<Integer> stack2 = new Stack<>();
public void push(int node) {
stack1.push(node);
if(stack2.isEmpty()){
stack2.push(node);
}else if(stack2.peek()>node){
stack2.push(node);
}
}
public void pop() {
if(stack1.peek()==stack2.peek()){
stack1.pop();
stack2.pop();
}else{
stack1.pop();
}
}
public int top() {
return stack1.peek();
}
public int min() {
return stack2.peek();
}
}
