文章目录

• problem Ⅰ
• • my solution 1 `stack`
  • my solution 2 `cnt`
• problem Ⅱ
• • my solution `wrong`
  • my solution `queue`
• problem Ⅲ
• • solution

problem Ⅰ

1249. Minimum Remove to Make Valid Parentheses
Given a string s of ‘(’ , ‘)’ and lowercase English characters.

Your task is to remove the minimum number of parentheses ( ‘(’ or ‘)’, in any positions ) so that the resulting parentheses string is valid and return any valid string.

Formally, a parentheses string is valid if and only if:

It is the empty string, contains only lowercase characters, or
It can be written as AB (A concatenated with B), where A and B are valid strings, or
It can be written as (A), where A is a valid string.

Example 1:

Example 2:

Example 3:

my solution 1 stack

class Solution {
public:
    string minRemoveToMakeValid(string s) {
        stack<int> stk;
        for(int i=0; i<s.size(); i++){
            if(s[i]=='(')
                stk.push(i);
            else if(s[i]==')'){
                if(!stk.empty())
                    stk.pop();
                else
                    s[i]='#';
            }
        }
        while(!stk.empty()){
            s[stk.top()]='#';
            stk.pop();
        }
        string ans = "";
        for(int i=0; i<s.size(); i++){
            if(s[i] != '#')
                ans.push_back(s[i]);
        }
        return ans;
    }
};

my solution 2 cnt

class Solution {
public:
    string minRemoveToMakeValid(string s) {
        int cnt = 0;
        // from left to right
        for(int i=0; i<s.size(); i++){
            if(s[i] == '(') cnt++;
            else if(s[i] == ')'){
                if(cnt == 0) s[i] = '#';
                else cnt--;
            }
        }
        // from right to left
        cnt = 0;
        for(int i=s.size(); i>=0; i--){
            if(s[i] == ')') cnt++;
            else if(s[i] == '('){
                if(cnt == 0) s[i] = '#';
                else cnt--;
            }
        }
        //return ans
        string ans = "";
        for(int i=0; i<s.size(); i++){
            if(s[i] != '#')
                ans.push_back(s[i]);
        }
        return ans;
    }
};

problem Ⅱ

my solution wrong

class Solution {
public:
    int findTheWinner(int n, int k) {
        vector<int> que(n, 0);
        int idx = 0;
        for(int i=0; i<n-1; i++){// n loop
            int step = 0;
            while(step < k-1){
                if(que[idx]==0)
                    step++;
                idx = (idx+1)%n;
            }
            que[idx]=1;
            idx = (idx+1)%n;
        }
        for(int i=0; i<n; i++){
            if(que[i]==0)return i+1;
        }
        return -1;
    }
};

my solution queue

class Solution {
public:
    int findTheWinner(int n, int k) {
        queue<int> que;
        for(int i=1; i<=n; i++)
            que.push(i);
        while(que.size() != 1){
            for(int i=0; i<k-1; i++){
                int tmp = que.front();
                que.pop();
                que.push(tmp);
            }
            que.pop();
        }
        return que.front();
    }
};

time complexity : $O(n)+O((n-1)\ast (k-1))=O(n\ast k)$
space complexity :$O(n)$

problem Ⅲ

155. Min Stack
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.

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

solution

class MinStack {
public:
    vector< pair<int,int> > s;
	
    MinStack() { }
    
    void push(int val) {
        if(s.empty())
            s.push_back({val,val});
        else
            s.push_back({val,min(s.back().second,val)});    
    }
    
    void pop() { s.pop_back(); }
    
    int top() { return s.back().first; }
    
    int getMin() { return s.back().second; }
};

/**
 * 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();
 */