leetcode155.MinStack

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.


将min设成long是因为可能是Integer.MAX_VALUE-Integer.MIN_VALUE
栈里存储的是push的值与最小值的差值。如果，差值小于零，说明push的值比最小值还要小，当取出的时候如果是小于零的值，取出的值就是min同时更新min

class MinStack {
long min;
Stack st;
/** initialize your data structure here. */
public MinStack() {
st=new Stack<>();
}

public void push(int x) {
if(st.isEmpty()){
st.push(0L);
min=x;
}
else{
st.push(x-min);
if(x min=x;
}
}

public void pop() {
if(st.isEmpty())return;
long p=st.pop();
if(p<0)
min=min-p;
}

public int top() {
long t=st.peek();
if(t<0)
return (int)min;
else
return (int)(t+min);
}

public int getMin() {
return (int)min;
}
}

这还有个绝妙的做法

class MinStack {
int min = Integer.MAX_VALUE;
Stack stack = new Stack();
public void push(int x) {
// only push the old minimum value when the current
// minimum value changes after pushing the new value x
if(x <= min){
stack.push(min);
min=x;
}
stack.push(x);
}
public void pop() {
// if pop operation could result in the changing of the current minimum value,
// pop twice and change the current minimum value to the last minimum value.
if(stack.pop() == min) min=stack.pop();
}
public int top() {
return stack.peek();
}
public int getMin() {
return min;
}
}