原题链接在这里:https://leetcode.com/problems/max-stack/description/
题目:
Design a max stack that supports push, pop, top, peekMax and popMax.
- push(x) -- Push element x onto stack.
- pop() -- Remove the element on top of the stack and return it.
- top() -- Get the element on the top.
- peekMax() -- Retrieve the maximum element in the stack.
- popMax() -- Retrieve the maximum element in the stack, and remove it. If you find more than one maximum elements, only remove the top-most one.
Example 1:
MaxStack stack = new MaxStack();
stack.push(5);
stack.push(1);
stack.push(5);
stack.top(); -> 5
stack.popMax(); -> 5
stack.top(); -> 1
stack.peekMax(); -> 5
stack.pop(); -> 1
stack.top(); -> 5
Note:
- -1e7 <= x <= 1e7
- Number of operations won't exceed 10000.
- The last four operations won't be called when stack is empty.
题解:
两个stack来解决. 不同的是popMax, 用temp stack 来保存找到最大值之前的, pop出max后再加回去, 同时更新maxStk.
Note: when popMax, get temp back, it needs to update maxStk as well.
Time Complexity: push, O(1). pop, O(1). top, O(1). peekMax, O(1). popMax, O(n). n是stack中数字的个数.
Space: O(n).
AC Java:
class MaxStack {
Stack<Integer> stk;
Stack<Integer> maxStk; /** initialize your data structure here. */
public MaxStack() {
stk = new Stack<Integer>();
maxStk = new Stack<Integer>();
} public void push(int x) {
stk.push(x);
if(maxStk.isEmpty() || maxStk.peek()<=x){
maxStk.push(x);
}
} public int pop() {
int x = stk.pop();
if(!maxStk.isEmpty() && x==maxStk.peek()){
maxStk.pop();
} return x;
} public int top() {
return stk.peek();
} public int peekMax() {
return maxStk.peek();
} public int popMax() {
Stack<Integer> tempStk = new Stack<Integer>();
int x = maxStk.pop();
while(!stk.isEmpty() && stk.peek()<x){
tempStk.push(stk.pop());
} stk.pop();
while(!tempStk.isEmpty()){
int top = tempStk.pop();
push(top);
}
return x;
}
} /**
* Your MaxStack object will be instantiated and called as such:
* MaxStack obj = new MaxStack();
* obj.push(x);
* int param_2 = obj.pop();
* int param_3 = obj.top();
* int param_4 = obj.peekMax();
* int param_5 = obj.popMax();
*/
类似Min Stack.