2678. 老人的数目（简单遍历模拟）

https://leetcode.cn/problems/number-of-senior-citizens/description/?envType=daily-question&envId=2023-10-23

class Solution {public int countSeniors(String[] details) {int ans = 0;for (String s: details) {int age = (s.charAt(11) - '0') * 10 + s.charAt(12) - '0';ans += age > 60? 1: 0;}return ans;}
}

会比下面的代码快一些。

class Solution {public int countSeniors(String[] details) {int ans = 0;for (String detail: details) {int age = Integer.parseInt(detail.substring(11, 13));ans += age > 60? 1: 0;}return ans;}
}

1155. 掷骰子等于目标和的方法数（动态规划）

https://leetcode.cn/problems/number-of-dice-rolls-with-target-sum/description/?envType=daily-question&envId=2023-10-24

提示：

1 <= n, k <= 30
1 <= target <= 1000

数据范围很小，采用三层循环。

class Solution {public int numRollsToTarget(int n, int k, int target) {long[][] dp = new long[n + 1][target + 1];final long MOD = (long)1e9 + 7;dp[0][0] = 1;for (int i = 1; i <= n; ++i) {          // 枚举骰子for (int j = 1; j <= k; j++) {      // 枚举当前面for (int x = 0; x <= target - j; ++x) { // 枚举上个骰子的和dp[i][x + j] = (dp[i][x + j] + dp[i - 1][x]) % MOD;}}}return (int)dp[n][target];}
}

2698. 求一个整数的惩罚数（预处理+dfs回溯）

https://leetcode.cn/problems/find-the-punishment-number-of-an-integer/description/?envType=daily-question&envId=2023-10-25

提示：
1 <= n <= 1000

class Solution {static int[] ans = new int[1001];static int target = 0;// 预处理static {for (int i = 1; i <= 1000; ++i) {if (op(i)) {ans[i] = ans[i - 1] + i * i;} else ans[i] = ans[i - 1];}}public int punishmentNumber(int n) {System.out.println(op(1));return ans[n];}// 判断x是否满足条件public static boolean op(int x) {String s = String.valueOf(x * x);target = x;return dfs(s, 0, 0);}public static boolean dfs(String s, int i, int t) {if (i == s.length() && t == target) return true;if (i >= s.length()) return false;boolean res = false;for (int j = i + 1; j <= s.length() && !res; ++j) {res |= dfs(s, j, t + Integer.parseInt(s.substring(i, j)));}return res;}
}

2520. 统计能整除数字的位数（简单模拟）

https://leetcode.cn/problems/count-the-digits-that-divide-a-number/description/?envType=daily-question&envId=2023-10-26

提示：
1 <= num <= 10^9
num 的数位中不含 0

class Solution {public int countDigits(int num) {int t = num, ans = 0;while (t != 0) {if (num % (t % 10) == 0) ans++;t /= 10;}return ans;}
}

1465. 切割后面积最大的蛋糕（贪心）

https://leetcode.cn/problems/maximum-area-of-a-piece-of-cake-after-horizontal-and-vertical-cuts/description/?envType=daily-question&envId=2023-10-27

贪心得想，任意两个长和宽都可以组合起来。那么最大面积就是由最大的长和宽组合起来的结果。

class Solution {public int maxArea(int h, int w, int[] horizontalCuts, int[] verticalCuts) {Arrays.sort(horizontalCuts);Arrays.sort(verticalCuts);int m = horizontalCuts.length, n = verticalCuts.length;int mxH = Math.max(h - horizontalCuts[m - 1], horizontalCuts[0]), mxW = Math.max(w - verticalCuts[n - 1], verticalCuts[0]);for (int i = 1; i < m; ++i) mxH = Math.max(mxH, horizontalCuts[i] - horizontalCuts[i - 1]);for (int i = 1; i < n; ++i) mxW = Math.max(mxW, verticalCuts[i] - verticalCuts[i - 1]);return (int)((long)mxH * mxW % (long)(1e9 + 7));}
}

2558. 从数量最多的堆取走礼物（优先队列）

https://leetcode.cn/problems/take-gifts-from-the-richest-pile/description/?envType=daily-question&envId=2023-10-28

提示：
1 <= gifts.length <= 10^3
1 <= gifts[i] <= 10^9
1 <= k <= 10^3

class Solution {public long pickGifts(int[] gifts, int k) {long s = 0;PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> b - a);for (int g: gifts) {s += g;pq.offer(g);}for (int i = 0 ; i < k; ++i) {int v = pq.poll(), x = (int)Math.sqrt(v);s -= v - x;pq.offer(x);}return s;}
}

274. H 指数（二分查找）

https://leetcode.cn/problems/h-index/description/?envType=daily-question&envId=2023-10-29

提示：

n == citations.length
1 <= n <= 5000
0 <= citations[i] <= 1000

先排序，再二分

class Solution {public int hIndex(int[] citations) {Arrays.sort(citations);int n = citations.length, l = 0, r = n; // 二分hwhile (l < r) {int mid = l + r + 1 >> 1, v = citations[n - mid];if (v >= mid) l = mid;else r = mid - 1;}return l;}
}

O(n)计数排序

见：https://leetcode.cn/problems/h-index/solutions/869042/h-zhi-shu-by-leetcode-solution-fnhl/

倒序枚举统计引用数量>=i的论文数量。

class Solution {public int hIndex(int[] citations) {int n = citations.length, tot = 0;int[] cnt = new int[n + 1];for (int i = 0; i < n; ++i) {cnt[Math.min(citations[i], n)]++;}for (int i = n; i >= 0; --i) {tot += cnt[i];if (tot >= i) return i;}return 0;}
}