有序三元组中的最大值 I

class Solution {
public:long long maximumTripletValue(vector<int>& nums) {vector<long long> num;for (auto &item:nums) {num.push_back(item*1ll);}long long z = 0,f = 1000000;long long ans = 0;long long maxx = num[0],minn = num[0];for (int i = 1;i < nums.size();i ++) {if (i > 1 ) {if (nums[i] > 0) {ans = max(ans,num[i]*z);} else {ans = max(ans,num[i]*f);}}z = max(z,maxx-num[i]);f = min(f,minn-num[i]);minn = min(minn,num[i]);maxx = max(maxx,num[i]);}return ans;}
};

有序三元组中的最大值 II

class Solution {
public:long long maximumTripletValue(vector<int>& nums) {vector<long long> num;for (auto &item:nums) {num.push_back(item*1ll);}long long z = 0,f = 1000000;long long ans = 0;long long maxx = num[0],minn = num[0];for (int i = 1;i < nums.size();i ++) {if (i > 1 ) {if (nums[i] > 0) {ans = max(ans,num[i]*z);} else {ans = max(ans,num[i]*f);}}z = max(z,maxx-num[i]);f = min(f,minn-num[i]);minn = min(minn,num[i]);maxx = max(maxx,num[i]);}return ans;}
};

无限数组的最短子数组

• 题意
  
• 思路
  把target/nums_sum 个数加上，剩下的就是拼接一下找最短子数组

class Solution {
public:int minSizeSubarray(vector<int> &nums, int target) {long long total = accumulate(nums.begin(), nums.end(), 0LL);int n = nums.size();int ans = INT_MAX;int left = 0;long long sum = 0;for (int right = 0; right < n * 2; right++) {sum += nums[right % n];while (sum > target % total) {sum -= nums[left++ % n];}if (sum == target % total) {ans = min(ans, right - left + 1);}}return ans == INT_MAX ? -1 : ans + target / total * n;}
};

有向图访问计数（待补）

• 题意
  
• 思路
  
• 代码

class Solution {
public:vector<int> countVisitedNodes(vector<int>& edges) {int n = edges.size();// e[sn] 表示从 sn 出发的反向边（即原图以 sn 为终点的边）vector<int> e[n];for (int i = 0; i < n; i++) e[edges[i]].push_back(i);// vis 数组用来求环长int vis[n];memset(vis, 0, sizeof(vis));// dis[i] 表示从点 i 走到环上的距离int dis[n];memset(dis, -1, sizeof(dis));vector<int> ans(n);// 对每个连通块分别求答案for (int S = 0; S < n; S++) if (dis[S] == -1) {// 从任意一点出发，都能走到环上，模拟求环长int now = S, cnt = 1;for (; vis[now] == 0; now = edges[now], cnt++) vis[now] = cnt;int loop = cnt - vis[now];// 从环上每个点开始，在反向边上 bfs，求每个点走到环上的距离queue<int> q;q.push(now); dis[now] = 0;for (int sn = edges[now]; sn != now; sn = edges[sn]) q.push(sn), dis[sn] = 0;while (!q.empty()) {int sn = q.front(); q.pop();// 答案就是环长 + 走到环上的距离ans[sn] = dis[sn] + loop;for (int fn : e[sn]) if (dis[fn] == -1) {q.push(fn);dis[fn] = dis[sn] + 1;}}}return ans;}
};