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

本文链接：https://blog.youkuaiyun.com/ShowMeTheCod3/article/details/127141628

本文介绍了两个具体的算法问题解决方案：计算两个整数的公因子数量及在二维网格中寻找具有最大总和的沙漏形状。通过详细的代码实现，展示了如何高效解决这些常见算法挑战。

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公因子的数目

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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

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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

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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

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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;
    }
};