模拟

class Solution {public:int scoreOfString(string s) {int res = 0;for (int i = 1; i < s.size(); i++)        res += abs(s[i] - s[i - 1]);return res;}
};

排序：先按照 $x_i$ 排序，然后顺序遍历数组，尽可能将 $points[i]$ 放入已有的矩形中

class Solution {public:int minRectanglesToCoverPoints(vector<vector<int>>& points, int w) {sort(points.begin(), points.end());int res = 0;int n = points.size();for (int i = 0, j = 0; i < n; i = ++j) {res++;while (j + 1 < n && points[j + 1][0] - points[i][0] <= w)j++;}return res;}
};

最短路: dijkstra+一个节点时间判断

class Solution {public:vector<int> minimumTime(int n, vector<vector<int>>& edges, vector<int>& disappear) {vector<pair<int, int>> e[n];for (auto& edge : edges) {e[edge[0]].push_back({edge[1], edge[2]});e[edge[1]].push_back({edge[0], edge[2]});}int inf = 1e9;vector<int> p(n, inf);p[0] = 0;priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> q;q.emplace(0, 0);while (!q.empty()) {auto [d, u] = q.top();q.pop();if (p[u] < d)continue;for (auto& [v, w] : e[u]) {int npv = d + w;if (npv < p[v] && npv < disappear[v]) {p[v] = npv;q.emplace(npv, v);}}}for (auto& i : p)if (i == inf)i = -1;return p;}
};

单调栈+哈希：遍历数组，维护一个非递增的栈，每次出栈时，将出栈元素的出现次数清零，每次元素进栈时出现次数+1，同时更新答案

class Solution {public:long long numberOfSubarrays(vector<int>& nums) {long long res = 0;stack<int> st;unordered_map<int, int> cnt;for (int i = 0; i < nums.size(); i++) {while (!st.empty() && st.top() < nums[i]) {cnt[st.top()] = 0;st.pop();}res += ++cnt[nums[i]];st.push(nums[i]);}return res;}
};