A leetcode.cn/problems/modify-the-matrix/description/" rel="nofollow">修改矩阵

模拟

class Solution {
public:
    vector<vector<int>> modifiedMatrix(vector<vector<int>> &matrix) {
        int m = matrix.size(), n = matrix[0].size();
        vector<int> mx(n, INT32_MIN);
        for (int i = 0; i < m; i++)
            for (int j = 0; j < n; j++)
                mx[j] = max(mx[j], matrix[i][j]);
        for (int i = 0; i < m; i++)
            for (int j = 0; j < n; j++)
                if (matrix[i][j] == -1)
                    matrix[i][j] = mx[j];
        return matrix;
    }
};

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B leetcode.cn/problems/number-of-subarrays-that-match-a-pattern-i/description/" rel="nofollow">匹配模式数组的子数组数目 I

枚举 $nums$ 中长为 $m+1$ 的子数组，判断是否匹配模式数组

class Solution {
public:
    int countMatchingSubarrays(vector<int> &nums, vector<int> &pattern) {
        int n = nums.size(), m = pattern.size();
        int res = 0;
        for (int i = 0, j = m; j < n; i++, j++) {
            int k = 0;
            for (; k < m; k++) {
                if (nums[k + i + 1] > nums[k + i] && pattern[k] != 1)
                    break;
                if (nums[k + i + 1] == nums[k + i] && pattern[k] != 0)
                    break;
                if (nums[k + i + 1] < nums[k + i] && pattern[k] != -1)
                    break;
            }
            if (k == m)
                res++;
        }
        return res;
    }
};

---

C leetcode.cn/problems/maximum-palindromes-after-operations/description/" rel="nofollow">回文字符串的最大数量

贪心：记录 $words$ 中各字符总数，得到形成配对的字符数目 $even$ 和单一的字符数目 $odd$，按 $words$ 中字符串长短升序枚举字符串，判断能否形成回文字符串

class Solution {
public:
    int maxPalindromesAfterOperations(vector<string> &words) {
        vector<int> cnt(26);
        vector<int> li;
        for (auto &s: words) {
            li.push_back(s.size());
            for (auto c: s)
                cnt[c - 'a']++;
        }
        sort(li.begin(), li.end());
        int odd = 0, even = 0;//even:形成配对的字符数目,odd:单一的字符数目
        for (int i = 0; i < 26; i++)
            if (cnt[i]) {
                even += cnt[i] / 2 * 2;
                odd += cnt[i] % 2;
            }
        int res = 0;
        for (auto x: li) {
            if (x & 1) {
                if (even < x - 1)
                    break;
                even -= x - 1;
                if (odd == 0) {
                    if (even == 0)
                        break;
                    //需要拆一对配对字符
                    even--;
                    odd++;
                } else
                    odd--;
            } else {
                if (even < x)
                    break;
                even -= x;
            }
            res++;
        }
        return res;
    }
};

---

D leetcode.cn/problems/number-of-subarrays-that-match-a-pattern-ii/description/" rel="nofollow">匹配模式数组的子数组数目 II

滚动哈希 + 枚举：枚举 $nums$ 中长为 $m+1$ 的子数组，用哈希判断是否匹配模式数组

class Solution {
public:
    int countMatchingSubarrays(vector<int> &nums, vector<int> &pattern) {
        int n = nums.size(), m = pattern.size();
        vector<int> li;
        for (int i = 0; i + 1 < n; i++) {
            if (nums[i + 1] > nums[i])
                li.push_back(1);
            else if (nums[i + 1] == nums[i])
                li.push_back(0);
            else
                li.push_back(-1);
        }
        srand(time(0));//随机种子
        int e = 2333 + rand() % 100, p = 1e9 + rand() % 100;
        shash h1(li, e, p), h2(pattern, e, p);
        int res = 0;
        for (int i = 0; i + m - 1 < li.size(); i++)
            if (h1(i, i + m - 1) == h2(0, m - 1))
                res++;
        return res;
    }

    class shash {//滚动哈希模板
    public:
        using ll = long long;
        vector<ll> pres;
        vector<ll> epow;
        ll e, p;

        shash(vector<int> &s, ll e, ll p) {
            int n = s.size();
            this->e = e;
            this->p = p;
            pres = vector<ll>(n + 1);
            epow = vector<ll>(n + 1);
            epow[0] = 1;
            for (int i = 0; i < n; i++) {
                pres[i + 1] = (pres[i] * e + s[i]) % p;
                epow[i + 1] = (epow[i] * e) % p;
            }
        }

        ll operator()(int l, int r) {
            ll res = (pres[r + 1] - pres[l] * epow[r - l + 1] % p) % p;
            return (res + p) % p;
        }
    };
};