155. Min Stack
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题目:
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.
思路:本题,可以用C++内部栈实现,于是剩下就是实现getMin(),在实现过程中为了防止time limit,这需要设置一个标志位,与一个存储变量,用来存储上一个最小值(
避免重复运算,这个还是有很大弊端,但是leetcode的测试集并没有测出来);于是为了解决第一种内置栈方法的不足,我们额外增加一个内部栈用来存储最小值,具体代码如下:
代码1(内置栈):
class MinStack {
public:
/** initialize your data structure here. */
MinStack() {
stack<int> stk;
}
stack<int> stk;
int flag = 0;
int store;
void push(int x) {
stk.push(x);
flag = 1;
}
void pop() {
stk.pop();
flag = 1;
}
int top() {
return stk.top();
}
int getMin() {
if(flag==0)
return store;
stack<int> res = stk;
int temp = res.top();
while(!res.empty())
{
if(temp>res.top())
temp = res.top();
res.pop();
}
store =temp;
flag=0;
return temp;
}
};
/**
* 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();
*/
代码2(两个栈):
class MinStack {
private:
stack<int> stk1;
stack<int> stk2;
public:
/** initialize your data structure here. */
MinStack() {
}
void push(int x) {
stk1.push(x);
if(stk2.empty() || x <= stk2.top())
stk2.push(x);
}
void pop() {
int top = stk1.top();
stk1.pop();
if(top == stk2.top())
{
stk2.pop();
}
}
int top() {
return stk1.top();
}
int getMin() {
return stk2.top();
}
};
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