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_155.java
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package com.fishercoder.solutions.firstthousand;
import java.util.Deque;
import java.util.LinkedList;
import java.util.Stack;
public class _155 {
public static class Solution1 {
public static class MinStack {
private Stack<Integer> stack;
private int min;
public MinStack() {
stack = new Stack();
min = Integer.MAX_VALUE;
}
public void push(int x) {
if (x <= min) {
/*All the trick happens here, we push the second minimum number onto the stack before we push the newer one,
* this way, when popping, we could always get the next minimum one in constant time.*/
stack.push(min);
min = x;
}
stack.push(x);
}
public void pop() {
/*if the value on the top of the stack happens to be the current minimum, we'll pop twice and change
* the current min value to be the last min value */
if (min == stack.pop()) {
min = stack.pop();
}
}
public int top() {
return stack.peek();
}
public int getMin() {
return min;
}
}
}
public static class Solution2 {
/*
* We could store a pair onto the stack: the first element in the pair is the value itself,
* the second element in the pair is the current minimum element so far seen on the stack.
*/
public static class MinStack {
Deque<int[]> stack;
public MinStack() {
stack = new LinkedList<>();
}
public void push(int val) {
if (!stack.isEmpty()) {
int[] last = stack.peekLast();
int currentMin = last[1];
if (val <= currentMin) {
stack.addLast(new int[] {val, val});
} else {
stack.addLast(new int[] {val, currentMin});
}
} else {
stack.addLast(new int[] {val, val});
}
}
public void pop() {
stack.pollLast();
}
public int top() {
return stack.peekLast()[0];
}
public int getMin() {
return stack.peekLast()[1];
}
}
}
}