和T3一起讲

思路：贪心

猜了一个结论

$$\sum _{j=0}^{j=i}arr[j]>arr[i+1]，满足此条件的最大i$$

先对$arr$排序，逆序找到第一个$i$即可

class Solution {public long largestPerimeter(int[] nums) {// 思维题long sum = 0;Arrays.sort(nums);for (int i = 0; i < nums.length; i++) {sum += nums[i];}// 逆序for (int i = nums.length - 1; i >= 2; i--) {sum -= nums[i];if (sum > nums[i]) {return sum + nums[i];}}return -1;}}

思路： 前后缀拆解 + 单调栈上二分

因为题目要求最左侧和最右侧都严格递增，所以需要预处理前后缀，保证严格递增

从左往右枚举每个点v

• check后缀是递增的
• 寻找前缀构建的单调栈，且结尾小于v的点，累加数量
• 如果当前值前缀是递增的，则加入单调栈

class Solution {public long incremovableSubarrayCount(int[] nums) {int n = nums.length;boolean[] pre = new boolean[n];boolean[] suf = new boolean[n];pre[0] = suf[n - 1] = true;for (int i = 1; i < n; i++) {pre[i] = pre[i - 1] && nums[i] > nums[i - 1];}for (int i = n - 2; i >= 0; i--) {suf[i] = suf[i + 1] && nums[i] < nums[i + 1];}long res = 0;// java可以用treemap来偷鸡单调栈monostackTreeMap<Integer, Integer> range = new TreeMap<>();range.put(-1, 1); // 哨兵for (int i = 0; i < n; i++) {int v = nums[i];if (suf[i]) {var ent = range.lowerEntry(v);if (ent != null) {// 删除的子数组必要有1个元素，所以要分类讨论if (ent.getValue() + 1 == i + 2) {res += ent.getValue() - 1;} else {res += ent.getValue();}}}if (pre[i]) {// 为啥要+2, 主要是为了统计方便range.put(v, i + 2);}}// 处理尾巴{var ent = range.lowerEntry(Integer.MAX_VALUE);if (ent != null) {if (ent.getValue() + 1 == n + 2) {res += ent.getValue() - 1;} else {res += ent.getValue();}}}return res;}}

思路: 启发式合并

有一个结论：

$$如果一个序列arr,arr[0],arr[1],...,,arr[n-2],arr[n-1],抽取其中3个数使其乘积最大$$

$$取arr的最小3个数，a_1,a_2,a_3(a_1\le a_2\le a_3)$$

$$最大的3个数,b_1,b_2,b_3(b_1\le b_2\le b_3)$$

$$乘积最大=max(a_1\ast a_2\ast a_3,a_1\ast a_2\ast b_3,a_1\ast b_2\ast b_3,b_1\ast b_2\ast b_3)$$

有了这个结论后，剩下的就好办了

每个子节点再往上传的时候，只需要保留3个最小数，3个最大数即可。

而这点，就扣合本题的思路

$$启发式合并$$

class Solution {int n;List<Integer>[]g;int[] cost;long[] res;List<Integer> []mins;List<Integer> []maxs;void dfs(int u, int fa) {List<Integer> tmpMin = new ArrayList<>();List<Integer> tmpMax = new ArrayList<>();tmpMin.add(cost[u]);tmpMax.add(cost[u]);for (int v: g[u]) {if (v == fa) continue;dfs(v, u);for (int tv: mins[v]) {tmpMin.add(tv);}for (int tv: maxs[v]) {tmpMax.add(tv);}}if (tmpMin.size() < 3) {res[u] = 1;} else {Collections.sort(tmpMin);Collections.sort(tmpMax);// 核心逻辑long ans = Long.MIN_VALUE / 10;long a1 = tmpMin.get(0), a2 = tmpMin.get(1), a3 = tmpMin.get(2);int nz = tmpMax.size();long a4 = tmpMax.get(nz - 3), a5 = tmpMax.get(nz - 2), a6 = tmpMax.get(nz - 1);ans = Math.max(ans, a1 * a2 * a3);ans = Math.max(ans, a1 * a2 * a6);ans = Math.max(ans, a1 * a5 * a6);ans = Math.max(ans, a4 * a5 * a6);if (ans < 0) {res[u] = 0;} else {res[u] = ans;}}// 保留3位，往上传for (int i = 0; i < 3 && i < tmpMin.size(); i++) {mins[u].add(tmpMin.get(i));}for (int i = tmpMax.size() - 1; i >= 0 && tmpMax.size() - i <= 3; i--) {maxs[u].add(tmpMax.get(i));}}public long[] placedCoins(int[][] edges, int[] cost) {n = cost.length;g = new List[n];this.cost = cost;Arrays.setAll(g, x->new ArrayList<>());for (int[] e: edges) {g[e[0]].add(e[1]);g[e[1]].add(e[0]);}mins = new List[n];Arrays.setAll(mins, x->new ArrayList<>());maxs = new List[n];Arrays.setAll(maxs, x->new ArrayList<>());res = new long[n];dfs(0, -1);return res;}}