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手机免费建网站软件,怎样自己创造网站,网页怎么设计好看,青岛百度优化lc2335用大根堆每次抓最多的两种水各装一杯装完剩一种水就直接把剩余杯数算成时间#xff0c;最快装满所有杯子class Solution { public:int fillCups(vectorint a) {priority_queueint q;for (int x : a) if (x) q.push(x);int t 0;while (q.size() 最快装满所有杯子class Solution {public:int fillCups(vectorint a) {priority_queueint q;for (int x : a) if (x) q.push(x);int t 0;while (q.size() 2) {int f q.top(); q.pop();int s q.top(); q.pop();f--; s--; t;if (f) q.push(f);if (s) q.push(s);}if (!q.empty()) t q.top();return t;}};lc1057 选自行车piii dist_pq模拟贪心pq优先处理最优候选typedef pairint, int PII;typedef pairint, PII PIII;class Solution {public:vectorint assignBikes(vectorvectorint workers, vectorvectorint bikes) {int n workers.size(), m bikes.size();priority_queuePIII, vectorPIII, greaterPIII heap;vectorint ans(n, -1);for(int i 0; i n; i) {for(int j 0; j m; j) {int dist abs(workers[i][0] - bikes[j][0]) abs(workers[i][1] - bikes[j][1]);heap.push({dist, {i, j}});//init}}vectorint remain(m, 1);int cnt 0;while(cnt n) {auto t heap.top();heap.pop();int i t.second.first, j t.second.second;if(remain[j] ans[i] -1) {ans[i] j;remain[j] 0;cnt; //choice}}return ans;}};有迪杰斯特拉“贪心优先队列”的感觉迪杰斯特拉是用小根堆每次选“当前最短路径”的节点这里是用小根堆每次选“当前距离最小的工人-单车对”两者都是通过优先队列优先处理“最优候选”再逐步确定最终结果核心思路是一致的。不过这个解法有个小问题它把所有工人-单车对都入堆时间复杂度是 O(nm\log nm)n 是工人数m 是单车数当 n、m 很大时会比较耗时~还有一种桶排序的tricks计算距离时外循环从小到大遍历worker内循环从小到大遍历bike然后依次添加到指定桶的末尾这样同一个桶距离相同的的工人自行车对一定是工人id小的在前若工人id相同的则自行车编号小的在前正好符合题意后面只需要线性遍历就可以了省掉了耗时的排序过程class Solution {public:vectorint assignBikes(vectorvectorint workers, vectorvectorint bikes) {int n workers.size(), m bikes.size();// 桶的下标是距离桶内存储 (工人id, 单车id)vectorvectorpairint, int buckets(2001);// 外循环遍历工人从小到大内循环遍历单车从小到大for (int i 0; i n; i) {for (int j 0; j m; j) {int dist abs(workers[i][0] - bikes[j][0]) abs(workers[i][1] - bikes[j][1]);buckets[dist].emplace_back(i, j);}}vectorint ans(n, -1);vectorbool bike_used(m, false);int cnt 0;// 按距离从小到大遍历桶for (int d 0; d 2000 cnt n; d) {// 遍历当前距离桶内的所有 (工人, 单车) 对for (auto p : buckets[d]) {int worker_id p.first;int bike_id p.second;// 工人未分配 且 单车未被用if (ans[worker_id] -1 !bike_used[bike_id]) {ans[worker_id] bike_id;bike_used[bike_id] true;cnt;if (cnt n) break;}}}return ans;}};lc1430bfs1.if (i n - 1)return !t-left !t-right;2.if (!hasMatch) return false; //cutclass Solution {public:bool isValidSequence(TreeNode* root, vectorint arr){int n arr.size();if (!root) return false;queueTreeNode* q;q.push(root);int i 0;while (q.size()){int sz q.size();bool hasMatch false;while (sz--){auto t q.front();q.pop();if (arr[i] t-val){hasMatch true;// i是最后一个元素时必须是叶子节点if (i n - 1)return !t-left !t-right;if (t-left) q.push(t-left);if (t-right) q.push(t-right);}}if (!hasMatch) return false; //cuti;}return false;}};