365天挑战LeetCode1000题——Day 102 Airwallex 空中云汇 & 力扣周赛 公因子的数目 沙漏的最大总和 最小 XOR 对字母串可执行的最大删除数

公因子的数目

代码实现

class Solution {
public:
    int commonFactors(int a, int b) {
        int cnt = 0;
        int maxNum = min(a, b);
        for (int i = 1; i <= maxNum; i++) {
            if (!(a % i) && !(b % i)) cnt++;
        }
        return cnt;
    }
};

沙漏的最大总和

代码实现（前缀和）

class Solution {
public:
    int maxSum(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        vector<vector<long long>> dp(m, vector<long long>(n));
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                dp[i][j] = grid[i][j];
            }
        }
        for (int i = 1; i < m; i++) {
            dp[i][0] += dp[i - 1][0];
        }
        for (int j = 1; j < n; j++) {
            dp[0][j] += dp[0][j - 1];
        }
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                dp[i][j] += dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1];
                // cout << i << " " << j << " " << dp[i][j] << endl;
            }
        }
        long long answer = 0;
        for (int i = 2; i < m; i++) {
            for (int j = 2; j < n; j++) {
                if (i == 2 && j == 2) {
                    answer = max(answer, dp[i][j] - grid[i - 1][j] - grid[i - 1][j - 2]);
                    // cout << dp[i][j] << " " << grid[i - 1][j] << " " << grid[i - 1][j - 2];
                    continue;
                }
                if (i == 2) {
                    answer = max(answer, dp[i][j] - dp[i][j - 3] - grid[i - 1][j] - grid[i - 1][j - 2]);
                    continue;
                }
                if (j == 2) {
                    answer = max(answer, dp[i][j] - dp[i - 3][j] - grid[i - 1][j] - grid[i - 1][j - 2]);
                    continue;
                }
                answer = max(answer, dp[i][j] - dp[i - 3][j] - dp[i][j - 3] + dp[i - 3][j - 3]
                             - grid[i - 1][j] - grid[i - 1][j - 2]);
            }
        }
        return answer;
    }
};

最小 XOR

代码实现

class Solution {
public:
    int minimizeXor(int num1, int num2) {
        int tmp1 = num1, tmp2 = num2;
        int cnt1 = 0, cnt2 = 0;
        while (tmp1) {
            if (tmp1 & 1) cnt1++;
            tmp1 = tmp1 >> 1;
        }
        while (tmp2) {
            if (tmp2 & 1) cnt2++;
            tmp2 = tmp2 >> 1;
        }
        // cout << cnt1 << " " << cnt2 << endl;
        int shiftLeft = 30;
        int ans = 0;
        while (cnt1-- && cnt2--) {
            while (((1 << shiftLeft) & num1) == 0) shiftLeft--;
            ans += (1 << shiftLeft);
            // cout << shiftLeft << endl;
            shiftLeft--;
        }
        if (cnt2 < 0) return ans;
        // cout << cnt2;
        shiftLeft = 0;
        while (cnt2--) {
            // cout << ((1 << shiftLeft) & num1) << endl;
            while (((1 << shiftLeft) & num1) != 0) shiftLeft++;
            ans += (1 << shiftLeft);
            // cout << shiftLeft << endl;
            shiftLeft++;
        }
        return ans;
    }
};

对字母串可执行的最大删除数

代码实现

class Solution {
private:
    unordered_map<string, int> mp;
public:
    int deleteString(string s) {
        int i;
        for (i = 0; i < s.size(); i++) {
            if (s[i] != s[0]) {
                break;
            }
        }
        if (i == s.size()) return s.size();
        if (s.empty()) return 0;
        if (mp.count(s)) return mp[s];
        int maxCnt = 1;
        for (int i = 1; i <= s.size() / 2; i++) {
            if (s.substr(0, i) == s.substr(i, i)) {
                maxCnt = max(maxCnt, deleteString(s.substr(i)) + 1);
            }
        }
        mp[s] = maxCnt;
        return maxCnt;
    }
};