第 92 场双周赛

1.分割圆的最少切割次数

分割圆的最少切割次数

Solution

class Solution {
public:
    int numberOfCuts(int n) {
        int tmp = 360 / n;
        if(n == 1) return 0;
        if(n % 2 == 0) return n / 2;
        return n;
    }
};

2.行和列中一和零的差值

行和列中一和零的差值

Solution

class Solution {
public:
    vector<vector<int>> onesMinusZeros(vector<vector<int>>& grid) {
        int n = grid.size(), m = grid[0].size();
        int rcnt[n][2], ccnt[m][2];
        memset(rcnt,0,sizeof(rcnt));
        memset(ccnt,0,sizeof(ccnt));
        for(int i = 0;i < n;i++){
            for(int j = 0;j < m;j++){
                rcnt[i][grid[i][j]]++;
                ccnt[j][grid[i][j]]++;
            }
        }
        vector<vector<int>> ans(n, vector<int>(m));
        for(int i = 0;i < n;i++){
            for(int j = 0;j < m;j++){
                ans[i][j] = rcnt[i][1] + ccnt[j][1] - rcnt[i][0] - ccnt[j][0];
            }
        }
        return ans;
    }
};

3.商店的最少代价

商店的最少代价

Solution

class Solution {
public:
    int bestClosingTime(string customers) {
        int n = customers.size();
        vector<int> p1(n+1, 0); // 关门
        vector<int> p2(n+1, 0); // 开门
        for(int i=n-1;i>=0;i--) {
            p1[i] = p1[i+1] + (customers[i] == 'Y');
        }
        for(int i=0;i<n;i++) {
            p2[i+1] = p2[i] + (customers[i] == 'N');
        }
        int idx, ans = INT_MAX;
        for(int i=0;i<=n;i++) {  // 第i小时关门，
            int cost = p1[i] + p2[i];
            if(cost < ans) {
                ans = cost;
                idx = i;
            }
        }
        return idx;
    }
};

4.统计回文子序列数目

统计回文子序列数目

Solution

class Solution {
    const long MOD = 1e9 + 7;
public:
    int countPalindromes(string s) {
        int suf[10]{}, suf2[10][10]{}, pre[10]{}, pre2[10][10]{};
        for (int i = s.length() - 1; i >= 0; --i) {
            char d = s[i] - '0';
            for (int j = 0; j < 10; ++j)
                suf2[d][j] += suf[j];
            ++suf[d];
        }

        long ans = 0L;
        for (char d : s) {
            d -= '0';
            --suf[d];
            for (int j = 0; j < 10; ++j)
                suf2[d][j] -= suf[j]; // 撤销
            for (int j = 0; j < 10; ++j)
                for (int k = 0; k < 10; ++k)
                    ans += (long) pre2[j][k] * suf2[j][k]; // 枚举所有字符组合
            for (int j = 0; j < 10; ++j)
                pre2[d][j] += pre[j];
            ++pre[d];
        }
        return ans % MOD;
    }
};