【题目】实现一个特殊的栈,在实现栈的基本功能上再实现一个实现返回栈中最小元素的操作。
【要求】
1,pop、push、getMin操作的时间复杂度都是O(1);
2,设计的栈类型额可以使用现成的栈结构。
第一种代码实现:
public class GetMinStack_1 {
private Stack<Integer> stackData;
private Stack<Integer> stackMin;
public GetMinStack_1(){
stackData = new Stack<Integer>();
stackMin = new Stack<Integer>();
}
public void push(int newNum){
if (stackMin.isEmpty()) {
stackMin.push(newNum);
}else if (newNum <= getMin()) {
stackMin.push(newNum);
}
stackData.push(newNum);
}
public int pop(){
if (stackData.isEmpty()) {
System.out.println("stack is empty");
return -1;
}
int value = stackData.pop();
if (value == getMin()) {
stackMin.pop();
}
return value;
}
private int getMin() {
// TODO Auto-generated method stub
if (stackMin.isEmpty()) {
System.out.println("stack is empty");
return -1;
}
return stackMin.peek();
}
public static void main(String[] args) {
// TODO Auto-generated method stub
GetMinStack_1 stack = new GetMinStack_1();
int[] testNum = {4,2,4,6,5,0,1,10};
for(int i:testNum){
stack.push(i);
}
for(int i = 0; i < testNum.length; i++){
System.out.println(stack.getMin()+" "+stack.pop());
}
}
}
第二种代码实现 :
public class GetMinStack_2 {
private Stack<Integer> stackData;
private Stack<Integer> stackMin;
public GetMinStack_2(){
stackData = new Stack<Integer>();
stackMin = new Stack<Integer>();
}
public void push(int newNum){
if (stackMin.isEmpty()) {
stackMin.push(newNum);
}
else if (newNum <= getMin()) {
stackMin.push(newNum);
}else {
stackMin.push(getMin());
}
stackData.push(newNum);
}
public int pop(){
if (stackData.isEmpty()) {
System.out.println("stack is empty");
return -1;
}
int value = stackData.pop();
stackMin.pop();
return value;
}
private int getMin() {
// TODO Auto-generated method stub
if (stackMin.isEmpty()) {
System.out.println("stack is empty");
return -1;
}
return stackMin.peek();
}
public static void main(String[] args) {
// TODO Auto-generated method stub
GetMinStack_2 stack = new GetMinStack_2();
int[] testNum = {4,2,4,6,5,0,1,10};
for(int i:testNum){
stack.push(i);
}
for(int i = 0; i < testNum.length; i++){
System.out.println(stack.getMin()+" "+stack.pop());
}
}
}