155 Min Stack
155. Min Stack
1. Question
Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
push(x) -- Push element x onto stack.
pop() -- Removes the element on top of the stack.
top() -- Get the top element.
getMin() -- Retrieve the minimum element in the stack.
Example:
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); --> Returns -3.
minStack.pop();
minStack.top(); --> Returns 0.
minStack.getMin(); --> Returns -2.
2. Implementation
思路:这题主要注意的是对于push()和pop()两种操作中,当我们发现min时该怎么办。当栈顶是min时,我们需要在pop了min后,可以立马知道当前min应该变为哪个数,所以我们的想法是每当我们发现当前的元素比min小时,我们立马将min压栈,然后更新min值。这样的好处是,当我们call pop()时,如果栈顶的元素等于min, 我们可以很快的知道min应该更新为哪个数
class MinStack {
int min;
Stack<Integer> stack;
/** initialize your data structure here. */
public MinStack() {
min = Integer.MAX_VALUE;
stack = new Stack<>();
}
public void push(int x) {
if (x <= min) {
stack.push(min);
min = x;
}
stack.push(x);
}
public void pop() {
if (stack.peek() == min) {
stack.pop();
min = stack.pop();
}
else {
stack.pop();
}
}
public int top() {
return stack.peek();
}
public int getMin() {
return 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.getMin();
*/
3. Time & Space Complexity
所有操作的时间复杂度都是O(1), 空间复杂度为O(n), n是minStack当前存储的元素个数
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