155. Min Stack

标签：最小   shell   new   ret   const   思路   time   opera   ons   

Problem:

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 1:

Input
["MinStack","push","push","push","getMin","pop","top","getMin"]
[[],[-2],[0],[-3],[],[],[],[]]

Output
[null,null,null,null,-3,null,0,-2]

Explanation
MinStack minStack = new MinStack();
minStack.push(-2);
minStack.push(0);
minStack.push(-3);
minStack.getMin(); // return -3
minStack.pop();
minStack.top();    // return 0
minStack.getMin(); // return -2

Constraints:

• Methods pop, top and getMin operations will always be called on non-empty stacks.

思路：

定义一个栈与保存最小值的栈。最小值栈与栈同时push和pop，添加值的时候，与最小值栈顶元素比较，若新值小则将新值压入最小值栈，否则将最小值栈顶的元素再次压栈。

Solution (C++):

stack<int> stk, min_stk;
/** initialize your data structure here. */
MinStack() {
    
}

void push(int x) {
    stk.push(x);
    if (min_stk.empty() || x < min_stk.top())
        min_stk.push(x);
    else  min_stk.push(min_stk.top());
}

void pop() {
    if (!stk.empty() && !min_stk.empty()) {
        stk.pop();
        min_stk.pop();
    }
}

int top() {
    if (stk.empty())  return -1;
    return stk.top();
}

int getMin() {
    if (min_stk.empty())  return -1;
    return min_stk.top();
}

性能：

Runtime: 40 ms??Memory Usage: 14.5 MB

思路：

Solution (C++):

性能：

Runtime: ms??Memory Usage: MB

155. Min Stack

标签：最小   shell   new   ret   const   思路   time   opera   ons   

原文地址：https://www.cnblogs.com/dysjtu1995/p/12748844.html