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.
一刷
题解：在stack中存入的不是x, 而是x-min
例如3，4，2, 2
存入0，1（min=3), -1(min=2), 0
如果当前pop出来的小于0，则恢复原先的min
public class MinStack {
Stack<Long> stack;
long min = 0;
/** initialize your data structure here. */
public MinStack() {
stack = new Stack<>();
}
public void push(int x) {
if(stack.isEmpty()){
stack.push(0L);
min = x;
}
else{
stack.push(x-min);
if(x<min){
min = x;
}
}
}
public void pop() {
long cur = stack.pop();
if(cur<0){
long min_ori = min - cur;
cur = cur + min_ori;
min = min_ori;
}
}
public int top() {
long peek = stack.peek();
if(peek>=0) return (int) (min+peek);
else{
long min_ori = min - peek;
return (int) (peek + min_ori);
}
}
public int getMin() {
return (int)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();
*/
