[leetCode]剑指 Offer 30. 包含min函数的栈
解法
这题的关键是设计一种数据结构，当栈中压入节点时可以同时记录当前栈的最小值
class MinStack {
class Node{
int val;
int min;
Node next;
public Node(int val, int min){
this.val = val;
this.min = min;
}
}
private LinkedList<Node> stack;
/** initialize your data structure here. */
public MinStack() {
stack = new LinkedList<>();
}
public void push(int x) {
if(stack.isEmpty()){
Node node = new Node(x,x);
stack.push(node);
}else{
Node top = stack.peek();
int min = Math.min(x,top.min);
stack.push(new Node(x,min));
}
}
public void pop() {
stack.pop();
}
public int top() {
return stack.peek().val;
}
public int min() {
return stack.peek().min;
}
}
/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(x);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.min();
*/
自己实现栈
class MinStack {
class Node{
int val;
int min;
Node next;
public Node(int val, int min){
this.val = val;
this.min = min;
}
}
private Node first;
/** initialize your data structure here. */
public MinStack() {
}
public void push(int x) {
Node oldfirst = first;
int min = x;
if(oldfirst != null){
min = Math.min(oldfirst.min,x);
}
first = new Node(x,min);
first.next = oldfirst;
}
public void pop() {
first = first.next;
}
public int top() {
return first.val;
}
public int min() {
return first.min;
}
}
/**
* Your MinStack object will be instantiated and called as such:
* MinStack obj = new MinStack();
* obj.push(x);
* obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.min();
*/
