LeetCode第258场周赛

心情

题目在此
忘记打了，补题中

第一题(反转单词前缀)

class Solution {
public:
    string reversePrefix(string w, char ch) {
        string res = "";
        int n = w.size();
        int id = -1;
        for (int i = 0; i < n; i ++){
            if (w[i]==ch){
                id = i;
                break;
            }
        }
        for (int i = id; i >= 0; i --)res += w[i];
        for (int i = id + 1; i < n; i ++)res += w[i];
        return res;
    }
};

第二题(可互换矩形的组数)

约分，排序，求和

typedef long long ll;
class Solution {
public:
    ll interchangeableRectangles(vector<vector<int>>& r) {
        ll res = 0;
        int n = r.size();
        for (int i = 0; i < n; i ++){
            int z = __gcd(r[i][0],r[i][1]);
            r[i][0] = r[i][0]/z;
            r[i][1] = r[i][1]/z;
        }
        sort(r.begin(),r.end(),[](vector<int> &a,vector<int> &b){
            if (a[0] == b[0])return a[1] < b[1];
            return a[0] < b[0];
        });
        ll num = 1;
        for (int i = 1; i < n; i ++){
            if (r[i][0]==r[i-1][0]&&r[i][1]==r[i-1][1])num++;
            else {
                res += num*(num-1)/2;
                num = 1;
            }
        }
        res += num*(num-1)/2;
        return res;
    }
};

第三题(两个回文子序列长度的最大乘积)

数据较小，直接暴力枚举。
枚举所有子序列，找出其中回文的，逐一比较。

class Solution {
public:
    bool jd(string a){	//判回文
        for (int i = 0,j = a.size() - 1; i <= j; i ++,j--){
            if (a[i]!=a[a.size()-i-1])return false;
        }
        return true;
    }
    bool jd1(int a,int b,int n){	//判是否每一位不一样
        for (int i = n-1; i >= 0; i --){
            int ta = a >> i & 1,tb = b >> i & 1;
            if (ta==1&&tb==1)return false;
        }
        return true;
    }
    int maxProduct(string s) {
        int n = s.size();
        int res = 1;
        vector<pair<string,int>> vs;
        map<string,int> mp;
        for (int i = 0; i < (1 << n); i ++){
            string a = "";
            for (int j = n-1; j >= 0; j --)
                if ((i >> j & 1) == 1)a += s[j];
            
            if (!jd(a))continue;
            vs.push_back(pair{a,i});
        }
        int le = vs.size();
        for (int i = 0; i < le; i ++){
            for (int j = i + 1; j < le; j ++){
                if (jd1(vs[i].second,vs[j].second,n)){
                    int tt = vs[i].first.size()*vs[j].first.size();
                    res = max(res,tt);
                }
            }
        }
        return res;
    }
};

第四题(每棵子树内缺失的最小基因值)

$待续$