警告
本文最后更新于 2022-10-07,文中内容可能已过时。
请设计一个栈,除了常规栈支持的pop与push函数以外,还支持min函数,该函数返回栈元素中的最小值。执行push、pop和min操作的时间复杂度必须为O(1)。
示例:
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| MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); --> 返回 -3.
minStack.pop();
minStack.top(); --> 返回 0.
minStack.getMin(); --> 返回 -2.
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题解
List、Stream 537ms
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| import java.util.ArrayList;
import java.util.Comparator;
import java.util.List;
class MinStack {
/** initialize your data structure here. */
List<Integer> stack = new ArrayList<>();
public MinStack() {
}
public void push(int x) {
stack.add(x);
}
public void pop() {
stack.remove(stack.size() -1);
}
public int top() {
return stack.get(stack.size() -1);
}
public int getMin() {
return stack.stream().min(Comparator.comparingInt(Integer::intValue)).get();
}
}
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官方 12ms
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| class MinStack {
/** initialize your data structure here. */
Deque<Integer> xStack;
Deque<Integer> minStack;
public MinStack() {
xStack = new LinkedList<Integer>();
minStack = new LinkedList<Integer>();
minStack.push(Integer.MAX_VALUE);
}
public void push(int x) {
xStack.push(x);
minStack.push(Math.min(minStack.peek(), x));
}
public void pop() {
xStack.pop();
minStack.pop();
}
public int top() {
return xStack.peek();
}
public int getMin() {
return minStack.peek();
}
}
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