文章目录

[0378. 有序矩阵中第K小的元素 [Medium] [Kth Smallest Element in a Sorted Matrix]](https://leetcode.com/problems/kth-smallest-element-in-a-sorted-matrix/)
[0441. 排列硬币 [Easy] [Arranging Coins]](https://leetcode.com/problems/arranging-coins/)
[0410. 分割数组的最大值 [Hard] [Split Array Largest Sum]](https://leetcode.com/problems/split-array-largest-sum/)
[0875. 爱吃香蕉的珂珂 [Medium] [Koko Eating Bananas]](https://leetcode.com/problems/koko-eating-bananas/)
[1011. 在 D 天内送达包裹的能力 [Medium] [Capacity To Ship Packages Within D Days]](https://leetcode.com/problems/capacity-to-ship-packages-within-d-days/)
[1283. 使结果不超过阈值的最小除数 [Medium] [Find the Smallest Divisor Given a Threshold]](https://leetcode.com/problems/find-the-smallest-divisor-given-a-threshold/)
[1760. 袋子里最少数目的球 [Medium] [Minimum Limit of Balls in a Bag]](https://leetcode.com/problems/minimum-limit-of-balls-in-a-bag/)
[1870. 准时到达的列车最小时速 [Medium] [Minimum Speed to Arrive on Time]](https://leetcode.com/problems/minimum-speed-to-arrive-on-time/)
[2064. 分配给商店的最多商品的最小值 [Medium] [Minimized Maximum of Products Distributed to Any Store]](https://leetcode.com/problems/minimized-maximum-of-products-distributed-to-any-store/)
[2187. 完成旅途的最少时间 [Medium] [Minimum Time to Complete Trips]](https://leetcode.com/problems/minimum-time-to-complete-trips/)

Input: matrix = [[1,5,9],[10,11,13],[12,13,15]], k = 8
Output: 13
Explanation: The elements in the matrix are [1,5,9,10,11,12,13,13,15], and the 8th smallest number is 13

Example 2:

Input: matrix = [[-5]], k = 1
Output: -5

Constraints:

Follow up:

题目：给你一个 n x n 矩阵 matrix ，其中每行和每列元素均按升序排序，找到矩阵中第 k 小的元素。请注意，它是 排序后 的第 k 小元素，而不是第 k 个不同的元素。你必须找到一个内存复杂度优于 O(n^2) 的解决方案。

**思路：**二分查找link，在[matrix[0][0], matrix[m-1][n-1] + 1)中二分查找目标值。

class Solution {public:int kthSmallest(vector<vector<int>>& matrix, int k) {int m = matrix.size();int n = matrix[0].size();int lo = matrix[0][0], hi = matrix[m-1][n-1] + 1;while(lo < hi){int mid = (lo + hi)/2;int cnt = 0, j = m-1;for(int i = 0; i < n; ++i){while(j >= 0 && matrix[i][j] > mid)--j;cnt += j + 1;}// 保证最后的lo一定为数组中的元素if(cnt < k)lo = mid + 1;elsehi = mid;}return lo;}
};

思路2：利用优先队列构建小顶堆。首先最小值肯定在matrix[0][0]，那么我们先将第一行放入堆中，堆顶为最小值，移除堆顶元素，同时将堆顶元素所在列的下一行对应的元素加入堆中。重复以上步骤k-1次，那么最后堆顶的即为第K小的元素。可参考Discuss。

工程代码下载

struct Tuple{public:int x;int y;int val;Tuple(int x, int y, int val):x(x), y(y), val(val){}
};struct CmpVal{bool operator() (const Tuple& t1, const Tuple& t2){return t1.val > t2.val;  // 小顶堆}
};class Solution {public:int kthSmallest(vector<vector<int>>& matrix, int k) {priority_queue<Tuple, vector<Tuple>, CmpVal> pq;int r = matrix.size();int c = r > 0 ? matrix[0].size() : 0;for(int j = 0; j < c; ++j)pq.push(Tuple(0, j, matrix[0][j]));for(int i = 1; i < k; ++i){Tuple t = pq.top();pq.pop();if(t.x >= r-1)continue;pq.push(Tuple(t.x + 1, t.y, matrix[t.x + 1][t.y]));}return pq.top().val;}
};

0441. 排列硬币 [Easy] [Arranging Coins]

You have n coins and you want to build a staircase with these coins. The staircase consists of k rows where the ith row has exactly i coins. The last row of the staircase may be incomplete.

Given the integer n, return the number of complete rows of the staircase you will build.

Example 1:

Input: n = 5
Output: 2
Explanation: Because the 3rd row is incomplete, we return 2.

Example 2:

Input: n = 8
Output: 3
Explanation: Because the 4th row is incomplete, we return 3.

Constraints:

1 <= n <= 2^31 - 1

题目：你总共有 n 枚硬币，并计划将它们按阶梯状排列。对于一个由 k 行组成的阶梯，其第 i 行必须正好有 i 枚硬币。阶梯的最后一行可能是不完整的。给你一个数字 n ，计算并返回可形成 完整阶梯行 的总行数。

思路：二分查找。

工程代码下载

class Solution {public:int arrangeCoins(int n) {long l = 1, r = (long)n + 1;int res = 0;while(l < r){long mid = l + (r - l)/2;long cnt = (1 + mid)*mid/2;if(cnt <= n){l = mid + 1;res = mid;}elser = mid;}return res;}
};

0410. 分割数组的最大值 [Hard] [Split Array Largest Sum]

Given an array nums which consists of non-negative integers and an integer m, you can split the array into m non-empty continuous subarrays.

Write an algorithm to minimize the largest sum among these m subarrays.

Example 1:

Input: nums = [7,2,5,10,8], m = 2
Output: 18
Explanation:
There are four ways to split nums into two subarrays.
The best way is to split it into [7,2,5] and [10,8],
where the largest sum among the two subarrays is only 18.

Example 2:

Input: nums = [1,2,3,4,5], m = 2
Output: 9

Example 3:

Input: nums = [1,4,4], m = 3
Output: 4

Constraints:

题目：给定一个非负整数数组 nums 和一个整数 m ，你需要将这个数组分成 m 个非空的连续子数组。设计一个算法使得这 m 个子数组各自和的最大值最小。

思路：二分法。假设最终求的结果为res，可知其在范围[max(nums), sum(nums)+1)中。当子串和的上限为mid时，计算其对应的子串数目cnt；与题目中对应的m比较，更改搜索范围。

工程代码下载

class Solution {public:int splitArray(vector<int>& nums, int m) {int lo = 0, hi = 1;for(int num : nums){lo = max(lo, num);hi += num;}while(lo < hi){int mid = lo + (hi-lo)/2;int cnt = 1, sum = 0;for(int num : nums){sum += num;if(sum > mid){cnt += 1;sum = num;}}if(cnt > m)lo = mid + 1;elsehi = mid;}return lo;}
};

0875. 爱吃香蕉的珂珂 [Medium] [Koko Eating Bananas]

Koko loves to eat bananas. There are n piles of bananas, the ith pile has piles[i] bananas. The guards have gone and will come back in h hours.

Koko can decide her bananas-per-hour eating speed of k. Each hour, she chooses some pile of bananas and eats k bananas from that pile. If the pile has less than k bananas, she eats all of them instead and will not eat any more bananas during this hour.

Koko likes to eat slowly but still wants to finish eating all the bananas before the guards return.

Return the minimum integer k such that she can eat all the bananas within h hours.

Example 1:

Input: piles = [3,6,7,11], h = 8
Output: 4

Example 2:

Input: piles = [30,11,23,4,20], h = 5
Output: 30

Example 3:

Input: piles = [30,11,23,4,20], h = 6
Output: 23

Constraints:

题目：珂珂喜欢吃香蕉。这里有 N 堆香蕉，第 i 堆中有 piles[i] 根香蕉。警卫已经离开了，将在 H 小时后回来。珂珂可以决定她吃香蕉的速度 K （单位：根/小时）。每个小时，她将会选择一堆香蕉，从中吃掉 K 根。如果这堆香蕉少于 K 根，她将吃掉这堆的所有香蕉，然后这一小时内不会再吃更多的香蕉。珂珂喜欢慢慢吃，但仍然想在警卫回来前吃掉所有的香蕉。返回她可以在 H 小时内吃掉所有香蕉的最小速度 K（K 为整数）。

思路：二分法，同1011. Capacity To Ship Packages Within D Days。

工程代码下载

class Solution {public:int minEatingSpeed(vector<int>& piles, int h) {int lo = 1, hi = 1e9 + 1;while(lo < hi){int mid = lo + (hi - lo) / 2;  // eating speedint cnt = 0;for(int p : piles){cnt += (p + mid - 1) / mid;}if(cnt > h)lo = mid + 1;elsehi = mid;}return lo;}
};

1011. 在 D 天内送达包裹的能力 [Medium] [Capacity To Ship Packages Within D Days]

A conveyor belt has packages that must be shipped from one port to another within D days.

The i-th package on the conveyor belt has a weight of weights[i]. Each day, we load the ship with packages on the conveyor belt (in the order given by weights). We may not load more weight than the maximum weight capacity of the ship.

Return the least weight capacity of the ship that will result in all the packages on the conveyor belt being shipped within D days.

Example 1:

Input: weights = [1,2,3,4,5,6,7,8,9,10], D = 5
Output: 15
Explanation:
A ship capacity of 15 is the minimum to ship all the packages in 5 days like this:
1st day: 1, 2, 3, 4, 5
2nd day: 6, 7
3rd day: 8
4th day: 9
5th day: 10Note that the cargo must be shipped in the order given, so using a ship of capacity 14 and splitting the packages into parts like (2, 3, 4, 5), (1, 6, 7), (8), (9), (10) is not allowed.

Example 2:

Input: weights = [3,2,2,4,1,4], D = 3
Output: 6
Explanation:
A ship capacity of 6 is the minimum to ship all the packages in 3 days like this:
1st day: 3, 2
2nd day: 2, 4
3rd day: 1, 4

Example 3:

Input: weights = [1,2,3,1,1], D = 4
Output: 3
Explanation:
1st day: 1
2nd day: 2
3rd day: 3
4th day: 1, 1

Note:

题目：传送带上的包裹必须在 days 天内从一个港口运送到另一个港口。传送带上的第 i 个包裹的重量为 weights[i]。每一天，我们都会按给出重量（weights）的顺序往传送带上装载包裹。我们装载的重量不会超过船的最大运载重量。返回能在 days 天内将传送带上的所有包裹送达的船的最低运载能力。

思路：同LeetCode410. Split Array Largest Sum

工程代码下载

class Solution {public:int shipWithinDays(vector<int>& weights, int days) {int lo = 1, hi = 1;for(int w : weights){lo = max(lo, w);hi += w;}while(lo < hi){int mid = (lo + hi) / 2;  // Capacityint d = 1, sum = 0;for(int w : weights){sum += w;if(sum > mid){sum = w;d += 1;}}if(d > days)lo = mid + 1;elsehi = mid;}return lo;}
};

1283. 使结果不超过阈值的最小除数 [Medium] [Find the Smallest Divisor Given a Threshold]

Given an array of integers nums and an integer threshold, we will choose a positive integer divisor, divide all the array by it, and sum the division’s result. Find the smallest divisor such that the result mentioned above is less than or equal to threshold.

Each result of the division is rounded to the nearest integer greater than or equal to that element. (For example: 7/3 = 3 and 10/2 = 5).

The test cases are generated so that there will be an answer.

Example 1:

Input: nums = [1,2,5,9], threshold = 6
Output: 5
Explanation: We can get a sum to 17 (1+2+5+9) if the divisor is 1.
If the divisor is 4 we can get a sum of 7 (1+1+2+3) and if the divisor is 5 the sum will be 5 (1+1+1+2).

Example 2:

Input: nums = [44,22,33,11,1], threshold = 5
Output: 44

Constraints:

题目：给你一个整数数组 nums 和一个正整数 threshold ，你需要选择一个正整数作为除数，然后将数组里每个数都除以它，并对除法结果求和。请你找出能够使上述结果小于等于阈值 threshold 的除数中最小的那个。

思路：二分法，参考lee215。更多二分查找题Link。

工程代码下载

class Solution {public:int smallestDivisor(vector<int>& nums, int threshold) {int lo = 1, hi = 1e6 + 1;while(lo < hi){int mid = (lo + hi) / 2;int sum = 0;for(int num : nums)sum += (num + mid - 1) / mid;if(sum > threshold)lo = mid + 1;elsehi = mid;}return lo;}
};

1760. 袋子里最少数目的球 [Medium] [Minimum Limit of Balls in a Bag]

You are given an integer array nums where the ith bag contains nums[i] balls. You are also given an integer maxOperations.

You can perform the following operation at most maxOperations times:

Take any bag of balls and divide it into two new bags with a positive number of balls.
- For example, a bag of 5 balls can become two new bags of 1 and 4 balls, or two new bags of 2 and 3 balls.

Your penalty is the maximum number of balls in a bag. You want to minimize your penalty after the operations.

Return the minimum possible penalty after performing the operations.

Example 1:

Input: nums = [9], maxOperations = 2
Output: 3
Explanation:
- Divide the bag with 9 balls into two bags of sizes 6 and 3. [9] -> [6,3].
- Divide the bag with 6 balls into two bags of sizes 3 and 3. [6,3] -> [3,3,3].
The bag with the most number of balls has 3 balls, so your penalty is 3 and you should return 3.

Example 2:

Input: nums = [2,4,8,2], maxOperations = 4
Output: 2
Explanation:
- Divide the bag with 8 balls into two bags of sizes 4 and 4. [2,4,8,2] -> [2,4,4,4,2].
- Divide the bag with 4 balls into two bags of sizes 2 and 2. [2,4,4,4,2] -> [2,2,2,4,4,2].
- Divide the bag with 4 balls into two bags of sizes 2 and 2. [2,2,2,4,4,2] -> [2,2,2,2,2,4,2].
- Divide the bag with 4 balls into two bags of sizes 2 and 2. [2,2,2,2,2,4,2] -> [2,2,2,2,2,2,2,2].
The bag with the most number of balls has 2 balls, so your penalty is 2 an you should return 2.

Example 3:

Input: nums = [7,17], maxOperations = 2
Output: 7

Constraints:

1 <= nums.length <= 10^5
1 <= maxOperations, nums[i] <= 10^9

题目：给你一个整数数组 nums ，其中 nums[i] 表示第 i 个袋子里球的数目。同时给你一个整数 maxOperations 。你可以进行如下操作至多 maxOperations 次：选择任意一个袋子，并将袋子里的球分到 2 个新的袋子中，每个袋子里都有 正整数 个球。比方说，一个袋子里有 5 个球，你可以把它们分到两个新袋子里，分别有 1 个和 4 个球，或者分别有 2 个和 3 个球。你的开销是单个袋子里球数目的 最大值 ，你想要 最小化 开销。请你返回进行上述操作后的最小开销。

思路：二分查找。假设我们已经找到了最小开销为res，那么对于某个袋子的球num，被分至n个新袋子，均满足个数小于等于res，其所需的拆分操作数为cnt=(num-1)/res。判断所有的操作数是否超过了maxOperations，依此来调整res。

工程代码下载

class Solution {public:int minimumSize(vector<int>& nums, int maxOperations) {int l = 1, r = 1e9;while(l < r){int mid = l + (r - l)/2;int cnt = 0;for(int num : nums){cnt += (num - 1) / mid;}if(cnt > maxOperations)l = mid + 1;elser = mid;}return l;}
};

1870. 准时到达的列车最小时速 [Medium] [Minimum Speed to Arrive on Time]

You are given a floating-point number hour, representing the amount of time you have to reach the office. To commute to the office, you must take n trains in sequential order. You are also given an integer array dist of length n, where dist[i] describes the distance (in kilometers) of the ith train ride.

Each train can only depart at an integer hour, so you may need to wait in between each train ride.

For example, if the 1st train ride takes 1.5 hours, you must wait for an additional 0.5 hours before you can depart on the 2nd train ride at the 2 hour mark.

Return the minimum positive integer speed (in kilometers per hour) that all the trains must travel at for you to reach the office on time, or -1 if it is impossible to be on time.

Tests are generated such that the answer will not exceed 10^7 and hour will have at most two digits after the decimal point.

Example 1:

Input: dist = [1,3,2], hour = 6
Output: 1
Explanation: At speed 1:
- The first train ride takes 1/1 = 1 hour.
- Since we are already at an integer hour, we depart immediately at the 1 hour mark. The second train takes 3/1 = 3 hours.
- Since we are already at an integer hour, we depart immediately at the 4 hour mark. The third train takes 2/1 = 2 hours.
- You will arrive at exactly the 6 hour mark.

Example 2:

Input: dist = [1,3,2], hour = 2.7
Output: 3
Explanation: At speed 3:
- The first train ride takes 1/3 = 0.33333 hours.
- Since we are not at an integer hour, we wait until the 1 hour mark to depart. The second train ride takes 3/3 = 1 hour.
- Since we are already at an integer hour, we depart immediately at the 2 hour mark. The third train takes 2/3 = 0.66667 hours.
- You will arrive at the 2.66667 hour mark.

Example 3:

Input: dist = [1,3,2], hour = 1.9
Output: -1
Explanation: It is impossible because the earliest the third train can depart is at the 2 hour mark.

Constraints:

题目：给你一个浮点数 hour ，表示你到达办公室可用的总通勤时间。要到达办公室，你必须按给定次序乘坐 n 趟列车。另给你一个长度为 n 的整数数组 dist ，其中 dist[i] 表示第 i 趟列车的行驶距离（单位是千米）。每趟列车均只能在整点发车，所以你可能需要在两趟列车之间等待一段时间。例如，第 1 趟列车需要 1.5 小时，那你必须再等待 0.5 小时，搭乘在第 2 小时发车的第 2 趟列车。返回能满足你准时到达办公室所要求全部列车的 最小正整数 时速（单位：千米每小时），如果无法准时到达，则返回 -1 。生成的测试用例保证答案不超过 10^7 ，且 hour 的 小数点后最多存在两位数字 。

思路：二分查找。由题目限制可知，可能的速度范围为[1, 10^7]。计算当前速度达到目的地所需的时间，二分查找对应的最小速度。更多二分查找题目:link。

工程代码下载

class Solution {public:int minSpeedOnTime(vector<int>& dist, double hour) {int l = 1, r = 1e7 + 1;int n= dist.size();while(l < r){int m = l + (r - l)/2;  // speeddouble t = 0;for(int i = 0; i < n-1; ++i){t += (dist[i] + m - 1) / m;}t += (double)dist.back() / m;if(t > hour)l = m + 1;elser = m;}return r == 1e7 + 1 ? -1 : r;}
};

2064. 分配给商店的最多商品的最小值 [Medium] [Minimized Maximum of Products Distributed to Any Store]

You are given an integer n indicating there are n specialty retail stores. There are m product types of varying amounts, which are given as a 0-indexed integer array quantities, where quantities[i] represents the number of products of the ith product type.

You need to distribute all products to the retail stores following these rules:

Return the minimum possible x.

Example 1:

Input: n = 6, quantities = [11,6]
Output: 3
Explanation: One optimal way is:
- The 11 products of type 0 are distributed to the first four stores in these amounts: 2, 3, 3, 3
- The 6 products of type 1 are distributed to the other two stores in these amounts: 3, 3
The maximum number of products given to any store is max(2, 3, 3, 3, 3, 3) = 3.

Example 2:

Input: n = 7, quantities = [15,10,10]
Output: 5
Explanation: One optimal way is:
- The 15 products of type 0 are distributed to the first three stores in these amounts: 5, 5, 5
- The 10 products of type 1 are distributed to the next two stores in these amounts: 5, 5
- The 10 products of type 2 are distributed to the last two stores in these amounts: 5, 5
The maximum number of products given to any store is max(5, 5, 5, 5, 5, 5, 5) = 5.

Example 3:

Input: n = 1, quantities = [100000]
Output: 100000
Explanation: The only optimal way is:
- The 100000 products of type 0 are distributed to the only store.
The maximum number of products given to any store is max(100000) = 100000.

Constraints:

题目：给你一个整数 n ，表示有 n 间零售商店。总共有 m 种产品，每种产品的数目用一个下标从 0 开始的整数数组 quantities 表示，其中 quantities[i] 表示第 i 种商品的数目。你需要将 所有商品 分配到零售商店，并遵守这些规则：一间商店至多只能有 一种商品 ，但一间商店拥有的商品数目可以为任意件。分配后，每间商店都会被分配一定数目的商品（可能为 0 件）。用 x 表示所有商店中分配商品数目的最大值，你希望 x 越小越好。也就是说，你想 最小化 分配给任意商店商品数目的 最大值 。请你返回最小的可能的 x 。

思路：二分查找。假设每个商店都能分到mid个商品，计算此时是否有足够的或多余的quantities，二分查找最小的mid。

工程代码下载

class Solution {public:int minimizedMaximum(int n, vector<int>& quantities) {int l = 1, r = 1e5 + 1;while(l < r){int mid = (l + r) / 2, sum = 0;for(int q : quantities)sum += (q + mid - 1) / mid;if(sum > n)l = mid + 1;elser = mid;}return l;}
};

2187. 完成旅途的最少时间 [Medium] [Minimum Time to Complete Trips]

You are given an array time where time[i] denotes the time taken by the ith bus to complete one trip.

Each bus can make multiple trips successively; that is, the next trip can start immediately after completing the current trip. Also, each bus operates independently; that is, the trips of one bus do not influence the trips of any other bus.

You are also given an integer totalTrips, which denotes the number of trips all buses should make in total. Return the minimum time required for all buses to complete at least totalTrips trips.

Example 1:

Input: time = [1,2,3], totalTrips = 5
Output: 3
Explanation:
- At time t = 1, the number of trips completed by each bus are [1,0,0].The total number of trips completed is 1 + 0 + 0 = 1.
- At time t = 2, the number of trips completed by each bus are [2,1,0].The total number of trips completed is 2 + 1 + 0 = 3.
- At time t = 3, the number of trips completed by each bus are [3,1,1].The total number of trips completed is 3 + 1 + 1 = 5.
So the minimum time needed for all buses to complete at least 5 trips is 3.

Example 2:

Input: time = [2], totalTrips = 1
Output: 2
Explanation:
There is only one bus, and it will complete its first trip at t = 2.
So the minimum time needed to complete 1 trip is 2.

Constraints:

题目：给你一个数组 time ，其中 time[i] 表示第 i 辆公交车完成 一趟****旅途所需要花费的时间。每辆公交车可以连续完成多趟旅途，也就是说，一辆公交车当前旅途完成后，可以立马开始下一趟旅途。每辆公交车独立运行，也就是说可以同时有多辆公交车在运行且互不影响。给你一个整数 totalTrips ，表示所有公交车总共需要完成的旅途数目。请你返回完成至少totalTrips 趟旅途需要花费的最少时间。

思路：二分查找。计算t时间内能够完成多少趟旅途，由限制条件可知，最坏的情况就是只有一辆车，对应的time为10^7、totalTrips为10^7，此时需要花费的时间为10^14，这个为花费时间的上限，依次进行二分查找。更多二分查找题目，参考link。

工程代码下载

class Solution {public:long long minimumTime(vector<int>& time, int totalTrips) {long long l = 1, r = 1e14 + 1;while(l < r){long long mid = l + (r -l)/2;long long cur = 0;for(int t : time)cur += mid / t;if(cur < totalTrips)l = mid + 1;elser = mid;}return l;}
};

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文章目录

0378. 有序矩阵中第K小的元素 [Medium] [Kth Smallest Element in a Sorted Matrix]

0441. 排列硬币 [Easy] [Arranging Coins]

0410. 分割数组的最大值 [Hard] [Split Array Largest Sum]

0875. 爱吃香蕉的珂珂 [Medium] [Koko Eating Bananas]

1011. 在 D 天内送达包裹的能力 [Medium] [Capacity To Ship Packages Within D Days]

1283. 使结果不超过阈值的最小除数 [Medium] [Find the Smallest Divisor Given a Threshold]

1760. 袋子里最少数目的球 [Medium] [Minimum Limit of Balls in a Bag]

1870. 准时到达的列车最小时速 [Medium] [Minimum Speed to Arrive on Time]

2064. 分配给商店的最多商品的最小值 [Medium] [Minimized Maximum of Products Distributed to Any Store]

2187. 完成旅途的最少时间 [Medium] [Minimum Time to Complete Trips]

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