模拟

class Solution {
public:vector<int> findIntersectionValues(vector<int> &nums1, vector<int> &nums2) {unordered_set<int> s1(nums1.begin(), nums1.end()), s2(nums2.begin(), nums2.end());vector<int> res(2);for (auto x: nums1)if (s2.count(x))res[0]++;for (auto x: nums2)if (s1.count(x))res[1]++;return res;}
};

动态规划：设 $p[i][j]$ 为将 $word[0,i]$ 修改为末位为 $j$ 的不含相邻近似相等字符串的最少操作数，枚举可能的 $word[i-1]$ 进行状态转移

class Solution {
public:int removeAlmostEqualCharacters(string word) {int n = word.size();int p[n][26];for (int j = 0; j < 26; j++)p[0][j] = 1;p[0][word[0] - 'a'] = 0;for (int i = 1; i < n; i++)for (int j = 0; j < 26; j++) {p[i][j] = INT32_MAX;for (int pre = 0; pre < 26; pre++)if (abs(j - pre) > 1)p[i][j] = min(p[i][j], p[i - 1][pre] + (word[i] - 'a' == j ? 0 : 1));}int res = INT32_MAX;for (int j = 0; j < 26; j++)res = min(res, p[n - 1][j]);return res;}
};

滑动窗口+哈希：哈希表记录滑动窗口内数的频率，枚举滑动窗口的左边界，尽可能移动滑动窗口的右边界

class Solution {
public:int maxSubarrayLength(vector<int> &nums, int k) {int n = nums.size();int res = 0;unordered_map<int, int> f;for (int l = 0, r = -1; l < n; f[nums[l++]]--) {while (r + 1 < n && f[nums[r + 1]] + 1 <= k)//滑窗左边界固定时，尽可能移动右边界f[nums[++r]]++;res = max(res, r - l + 1);}return res;}
};

枚举：枚举可能的关闭分部集合，然后在当前关闭情况下跑多源最短路算法，然后判断最大的最短路是否不超过 $maxDistance$

class Solution {
public:inline int get_bit(int mask, int loc) { return mask >> loc & 1; }//返回mask二进制表示的第loc位int pop_cnt(int mask) {//返回mask二进制表示中1的个数int res = 0;for (; mask; mask >>= 1)if (mask & 1)res++;return res;}int numberOfSets(int n, int maxDistance, vector<vector<int>> &roads) {int inf = INT32_MAX;vector<vector<int>> g(n, vector<int>(n, inf));for (auto &e: roads) {g[e[0]][e[1]] = min(g[e[0]][e[1]], e[2]);g[e[1]][e[0]] = g[e[0]][e[1]];}int res = 0;for (int mask = 0; mask < (1 << n); mask++) {//枚举关闭分部集合:mask二进制中第i位为0表示第i个分部关闭auto t = g;for (int k = 0; k < n; k++)if (get_bit(mask, k))for (int i = 0; i < n; i++)if (get_bit(mask, i))for (int j = 0; j < n; j++)if (get_bit(mask, j))if (t[i][k] != inf && t[k][j] != inf)t[i][j] = min(t[i][j], t[i][k] + t[k][j]);int mx = 0;//最大的最短路for (int i = 0; i < n; i++)if (get_bit(mask, i))for (int j = 0; j < n; j++)if (j != i && get_bit(mask, j))mx = max(mx, t[i][j]);if (mx <= maxDistance)res++;}return res;}
};