1.修改矩阵（模拟）

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

2.匹配模式数组的子数组数目 I（暴力）

class Solution {
public:
    int countMatchingSubarrays(vector<int>& nums, vector<int>& pattern) {
        int n = nums.size();
        int m = pattern.size();
        int ans = 0;
        //枚举nums
        for(int i = 0; i < n - m; i++){
            int f = 1;
            //枚举pattern
            for(int j = 0; j < m; j++){
            	如果不满足，那就跳到下一个
                if(nums[i + j + 1] > nums[i + j] && pattern[j] != 1) f = 0;
                if(nums[i + j + 1] == nums[i + j] && pattern[j] != 0) f = 0;
                if(nums[i + j + 1] < nums[i + j] && pattern[j] != -1) f = 0;
            }
            if(f) ans++;
        }
        return ans;
    }
};

3.回文字符串的最大数量（贪心）

class Solution {
public:
    int maxPalindromesAfterOperations(vector<string>& words) {
        int n = words.size();
        vector<int> cnt(26, 0);
        for(int i = 0; i < n; i++)
            for(char c : words[i])
                //统计每个字符有多少个
                cnt[c - 'a']++;
        int sum = 0, ans = 0;
        //统计有多少对字符
        for(int i : cnt) sum += i / 2;
        //从小到大排序，允许重复数字
        multiset<int> s;
        //插入每个字符串的长度
        for(int i = 0; i < n; i++) s.insert(words[i].size());
        for(int i : s){
            //计算剩余多少对字符
            int t = sum - i / 2;
            //如果剩余的对数大于等于0，那就更新答案和对数
            if(t >= 0){
                sum = t;
                ans++;
            }
        }
        return ans;
    }
};

4.匹配模式数组的子数组数目 II（KMP）

class Solution {
public:
    int countMatchingSubarrays(vector<int>& nums, vector<int>& pattern) {
        int n = nums.size();
        int m = pattern.size();
        vector<int> s(n), ne(n);
        //将匹配串转化为01串，用kmp匹配
        for(int i = 1; i < n; i++)
            s[i - 1] = nums[i] > nums[i - 1] ? 1 : nums[i] == nums[i - 1] ? 0 : -1;
        //求next模板串
        ne[0] = -1;
        for(int i = 1, j = -1; i < m; i++){
            while(j >= 0 && pattern[i] != pattern[j + 1]) j = ne[j];
            if(pattern[i] == pattern[j + 1]) j++;
            ne[i] = j;
        }
        int ans = 0;
        //kmp匹配，但是s里只有n-1个元素
        for(int i = 0, j = -1; i < n - 1; i++){
            while(j >= 0 && s[i] != pattern[j + 1]) j = ne[j];
            if(s[i] == pattern[j + 1]) j++;
            if(j == m - 1){
                ans++;
                j = ne[j];
            }
        }
        return ans;
    }
};