周 赛 补 题

力扣第320场周赛

竞赛 - 力扣 (LeetCode)备战技术面试？力扣提供海量技术面试资源，帮助你高效提升编程技能，轻松拿下世界 IT 名企 Dream Offer。https://leetcode.cn/contest/weekly-contest-320/1.数组中不等三元组的数目力扣https://leetcode.cn/problems/number-of-unequal-triplets-in-array/

只想到了暴力，三次循环一下满足

class Solution {
public:
    int unequalTriplets(vector<int>& nums) {
        int i=0;
        int ans=0;
        while(i<nums.size()-2){
            int j=i+1;
            while(j<nums.size()-1){
                int k=j+1;
                while(k<nums.size()){
                    if(nums[i] != nums[j]&&nums[i] != nums[k] && nums[j]!= nums[k])ans++;
                    k++;
                }
                j++;      
            }
          i++;
        }
        return ans;
    }
};

 2.二叉搜索树最近节点查询力扣https://leetcode.cn/problems/closest-nodes-queries-in-a-binary-search-tree/

先取出元素后通过二分来查找位置

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int>vec;
    void inorder(TreeNode* root)
    {
        if(root==nullptr)return;
        inorder(root->left);
        vec.push_back(root->val);
        inorder(root->right); 
    }
    tuple<int,int> find_pos(vector<int>& nums,int target)
    {
        if(target<nums[0])return {-1,nums[0]};
        else if(target>nums[nums.size()-1])return {nums[nums.size()-1],-1};
        int left=0;
        int right=nums.size()-1;
        int mid=(left+right)/2;
        while(left<=right)
        {
           if(nums[mid]==target)return {nums[mid],nums[mid]};
           else if(nums[mid]>target) right=mid-1;
           else left=mid+1;
           mid=(left+right)/2;
        }
        return {nums[right],nums[left]}; 
    }
    vector<vector<int>> closestNodes(TreeNode* root, vector<int>& queries) {
    vector<vector<int>> ans;
    inorder(root);
    for(auto i:queries)
    {
        vector<int>temp(2);
        auto[val1,val2]=find_pos(vec,i);
        temp[0]=val1;
        temp[1]=val2;
        ans.push_back(move(temp));
    }
    return ans;
    }
};

 ACwing第78场周赛

1.4719. 商品种类 - AcWing题库

商品名称和产地名字长度都不会超过10，所以分别把两者的长度用空格补充为10，然后拼接起来。之后插入set中，保证唯一性

#include <bits/stdc++.h>
using namespace std;

int main(){

    int n; 
    cin >> n;
    unordered_set<string> st;
    while(n--){
        string name, place;
        cin >> name >> place;
        while(name.size()<10) name = name + ' ';
        while(place.size()<10) place = place + ' ';
        st.insert(name + place);
    }
    cout<<st.size()<<endl;


    return 0;
}

2.4720. 字符串 - AcWing题库

开一个空栈来维护，string s的元素逐次进栈，只要接下来想进栈字符与top()相同，则说明是连续的字符，那么不但不能进栈，还要把top()给弹栈，最后剩下的就是没被消过的字符 

#include <iostream>
#include <cstring>
#include <algorithm>
#include <cstdio>
#include <string>
#include <stack>
using namespace std;
stack<char> stk;
string ans;

string removeDuplicates(string s) {
        for(auto item:s)
        {
            if(!stk.empty()&&stk.top()==item)
            stk.pop();
            else
            stk.push(item);
        }
        while(!stk.empty())
        {
            ans+=stk.top();
            stk.pop();
        }
        reverse(ans.begin(),ans.end());
        return ans;
    }
int main()
{
    string n;
    cin>>n;
    cout<<removeDuplicates(n);
    return 0;

}