标签:
Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.
class MinStack { LinkedList<Integer> stack = new LinkedList<>(); Integer min = Integer.MAX_VALUE; public void push(int x) { stack.push(x); if (x < min) { min = x; } } public void pop() { stack.pop(); min = Integer.MAX_VALUE; for (Integer index : stack) { if (index < min) { min = index; } } } public int top() { return stack.peek(); } public int getMin() { return min; } }
LinkedList<Integer> stack = new LinkedList<Integer>(); LinkedList<Integer> minStack = new LinkedList<Integer>(); int min = Integer.MAX_VALUE; public void push(int x) { stack.push(x); minStack.push(min); } public void pop() { stack.pop(); minStack.pop(); } public int top() { return stack.peek(); } public int getMin() { return minStack.peek(); }
class MinStack { LinkedList<Integer> stack = new LinkedList<Integer>(); LinkedList<Integer> minStack = new LinkedList<Integer>(); int min = Integer.MAX_VALUE; public void push(int x) { stack.push(x); if (min >= x) { min = x; minStack.push(min); } } public void pop() { if (min == stack.peek()) { minStack.pop(); if (minStack.isEmpty()) { min = Integer.MAX_VALUE; } else { min = minStack.peek(); } } stack.pop(); } public int top() { return stack.peek(); } public int getMin() { return min; } }
LinkedList<Long> stack = new LinkedList<>(); long min = Integer.MAX_VALUE; 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 pop = stack.pop(); if (pop < 0) { min -= pop; } } public int top() { long top = stack.peek(); if (top > 0) { return (int) (top+min); } else { return (int) min; } } public int getMin() { return (int) min; }运行效率最好的方法!!!
标签:
原文地址:http://blog.csdn.net/my_jobs/article/details/43525599