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2022 iThome 鐵人賽

DAY 22
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30天 Neetcode解題之路系列 第 22

Day 22 - 20. Valid Parentheses (By C++)

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前言

大家好,我是毛毛。ヾ(´∀ ˋ)ノ
那就開始今天的解題吧~


Question

Design a stack that supports push, pop, top, and retrieving the minimum element in constant time.

Implement the MinStack class:

MinStack() initializes the stack object.
void push(int val) pushes the element val onto the stack.
void pop() removes the element on the top of the stack.
int top() gets the top element of the stack.
int getMin() retrieves the minimum element in the stack.
You must implement a solution with O(1) time complexity for each function.

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:

  • -2^31 <= val <= 2^31 - 1
  • Methods pop, top and getMin operations will always be called on non-empty stacks.
  • At most 3 * 10^4 calls will be made to push, pop, top, and getMin.

Think

基本上就是要用陣列實作出stack的各種操作~

  • push
    • val塞進stack中
  • pop
    • 把stack最後一個放進去的值刪掉
  • getMin
    • 找出stack中最小值
  • top
    • 回傳目前最後一個放進去的值

Code

class MinStack {
public:
    int stack[30000];
    int ptr = -1;
    // int min_val;
    MinStack() {
        
    }
    
    void push(int val) {
        ptr++;
        stack[ptr] = val;
        // cout << "PUSH=> ptr: " << ptr << ", stack[ptr]: " << stack[ptr] << endl;
        
        // min_val = min_val < val? min_val : val;
        // cout << "min_val: " << min_val << endl;
    }
    
    void pop() {
        // cout << "B POP=> ptr: " << ptr << ", stack[ptr]: " << stack[ptr] << endl;
        ptr--;
        // cout << "A POP=> ptr: " << ptr << ", stack[ptr]: " << stack[ptr] << endl;
    }
    
    int top() {
        if(ptr == -1){
            return 0;
        } else {
            // cout << "TOP=> stack[ptr]: " << stack[ptr] << endl;
            return stack[ptr];
        }
        
    }
    
    int getMin() {
        int min_val;
        for (int index=0 ; index<=ptr ; index++){
            if(index == 0){
                min_val = stack[index];
            }
            if(min_val <= stack[index]){
                continue;
            } else {
                min_val = stack[index];
            }
        }
        return min_val;
    }
};

/**
 * Your MinStack object will be instantiated and called as such:
 * MinStack* obj = new MinStack();
 * obj->push(val);
 * obj->pop();
 * int param_3 = obj->top();
 * int param_4 = obj->getMin();
 */

Submission


今天就到這邊啦~
大家明天見/images/emoticon/emoticon29.gif


上一篇
Day 21 - 20. Valid Parentheses
下一篇
Day 23 - 150. Evaluate Reverse Polish Notation
系列文
30天 Neetcode解題之路30
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