2019-10-14 22:13:18 问题描述: 问题求解: 解法一:动态规划 这种数组划分的题目基本都可以使用dp来解决,核心的思路就是先维护低的划分,再在中间找分割点加入新的划分. public int splitArray(int[] nums, int m) { int n = nums.length; long[][] dp = new long[m + 1][n]; long[] presum = new long[n]; presum[0] = nums[0]; for (int…

［抄题］: Given an array 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. Note:If n is the length of array, as…

题意大概就是,给定一个包含非负整数的序列nums以及一个整数m,要求把序列nums分成m份,并且要让这m个子序列各自的和的最大值最小(minimize the largest sum among these m subarrays). Note:If n is the length of array, assume the following constraints are satisfied: 1 ≤ n ≤ 10001 ≤ m ≤ min(50, n) Examples: Input:num…

Given an array 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. Note:Given m satisfies the following const…

Given an array 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. Note:If n is the length of array, assume t…

Given an array 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. Note:Given m satisfies the following const…

Given an array 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. Note: Given m satisfies the following cons…

题目如下: Given an array 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 msubarrays. Note:If n is the length of array, ass…

Given an array 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. Note: Given m satisfies the following cons…

Given an array 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. Examples: Input: nums = [7,2,5,10,8] m = 2…

做了Zenefits的OA,比面经里的简单多了..害我担心好久 阴险的Baidu啊,完全没想到用二分,一开始感觉要用DP,类似于极小极大值的做法. 然后看了答案也写了他妈好久. 思路是再不看M的情况下,最终结果的取值范围是[最大的元素,所有元素之和],然后就是用二分在这个范围里找. 注意二分的取舍要配合取舍方程,我用的方程是,保证子集的和都小于等于二分测试的那个二分M,然后集合数小于要求的m.一旦子集数量超过m,说明我们测试的二分M太小了,弄大点,才能让每个子集多放点元素,从而使得总子集数不超过…

给定一个非负整数数组和一个整数 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.注意:数组长度 n 满足以下条件:    1 ≤ n ≤ 1000    1 ≤ m ≤ min(50, n)示例:输入:nums = [7,2,5,10,8]m = 2输出:18解释:一共有四种方法将nums分割为2个子数组.其中最好的方式是将其分为[7,2,5] 和 [10,8],因为此时这两个子数组各自的和的最大值为18,在所有情况中最小.详见:https:…

思路: dp. 实现: class Solution { public: int splitArray(vector<int>& nums, int m) { int n = nums.size(); vector<, ); ; i <= n; i++) sum[i] = sum[i - ] + nums[i - ]; vector<vector<, vector<, INT_MAX)); ; i <= n; i++) dp[i][] = sum[i…

You are given an integer array sorted in ascending order (may contain duplicates), you need to split them into several subsequences, where each subsequences consist of at least 3 consecutive integers. Return whether you can make such a split. Example…

In a given integer array A, we must move every element of A to either list B or list C. (B and C initially start empty.) Return true if and only if after such a move, it is possible that the average value of B is equal to the average value of C, and…

In a given integer array A, we must move every element of A to either list B or list C. (B and C initially start empty.) Return true if and only if after such a move, it is possible that the average value of B is equal to the average value of C, and…

Given an array with n integers, you need to find if there are triplets (i, j, k) which satisfies following conditions: 0 < i, i + 1 < j, j + 1 < k < n - 1 Sum of subarrays (0, i - 1), (i + 1, j - 1), (j + 1, k - 1) and (k + 1, n - 1) should be…

Given a string S of digits, such as S = "123456579", we can split it into a Fibonacci-like sequence [123, 456, 579]. Formally, a Fibonacci-like sequence is a list F of non-negative integers such that: 0 <= F[i] <= 2^31 - 1, (that is, each…

leetcode659. Split Array into Consecutive Subsequences 题意: 您将获得按升序排列的整数数组(可能包含重复项),您需要将它们拆分成多个子序列,其中每个子序列至少包含3个连续的整数.返回是否可以进行这样的拆分. 思路: 虽然是medium,但是我感觉有点难想= =. O(n)复杂度. dp. 先分开各种不同的序列.按如果间隔大于1就坑定是不同的序列.按此分开不同的序列以此简化问题. 然后处理连续的序列.连续的序列false的情况就是没有足够的数…

[LeetCode]659. Split Array into Consecutive Subsequences 解题报告(Python) 标签(空格分隔): LeetCode 作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址:https://leetcode.com/problems/split-array-into-consecutive-subsequences/description/ 题目描述: You are g…

/* 终于解决了IE8不支持数组的indexOf方法 */ if (!Array.prototype.indexOf) { Array.prototype.indexOf = function (elt /*, from*/) { var len = this.length >>> 0; var from = Number(arguments[1]) || 0; from = (from < 0) ? Math.ceil(from) : Math.floor(from); if (…

取出数组中最大值或最小值是开发中常见的需求,今天继续讲解如何获取javascript数组中最大和最小值. 1.排序法 首先我们给数组进行排序,可以按照从小到大的顺序来排,排序之后的数组中第一个和最后一个就是我们想要获取的最小值和最大值. 排序我们会用到数组的 sort 方法. var arr = [12,56,25,5,82,51,22]; arr.sort(function (a, b) { return a-b; }); // [5,12,22,25,51,56] var min = arr…

要求: 要求数组从文件读取. 如果输入的数组很大,  并且有很多大的数字,  就会产生比较大的结果 (考虑一下数的溢出), 请保证你的程序能正常输出. 另外, 如果输入文件的参数有错误, 这个程序应该能正常退出, 并显示相应的错误信息. 任何输入错误都不能导致你的程序崩溃. 解决方法及步骤: 1,先写入一个txt文件,再读取出来 2,通过循环测试生成的最大容量的数. 3,写入文件时控制数字的大小,就不会产生大数情况. 4,再加上上次的代码,计算最大连续子数组的和 代码: 这是随机生成一定量的数并…

javascript的内置对象Array是用于构造数组的全局对象,数组是类似于列表的高阶对象. 创建数组的方法: 1通过字面量:var arr = [1,2,3]; 里面的参数直接作为数组里的值 2通过构造器:var arr = new Array(1,2,3,4,5,6);  var arr2 = new Array(6);当构造器里的参数为一个时,作为数组实例的length属性. 原理:数组是一种类列表对象,它的原型中提供了遍历和修改元素的操作.javascript素组的长度和元素类型都是非…

659. Split Array into Consecutive Subsequences You are given an integer array sorted in ascending order (may contain duplicates), you need to split them into several subsequences, where each subsequences consist of at least 3 consecutive integers. Re…

ylbtech-Java-Runoob-高级教程-实例-数组:06. Java 实例 – 数组获取最大和最小值 1.返回顶部 1. Java 实例 - 数组获取最大和最小值  Java 实例 以下实例演示了如何通过 Collections 类的 Collections.max() 和 Collections.min() 方法来查找数组中的最大和最小值: Main.java 文件 import java.util.Arrays; import java.util.Collections; publ…

 重排数组求最大和 序号:#34难度:困难时间限制:1000ms内存限制:10M 描述 假设有一个n元素的数组(数组的元素索引从1开始),针对这个数组有q个查询请求,每个请求由一对整数li,ri组成,表示数组上一个合法索引区间(1≤li≤ri≤n).我们可以根据查询的索引来计算出范围内的所有数组元素的一个总和,计算时也需要包含li和ri位置的元素. 通过重新排序数组元素,可以使得这些查询的总和最大.求问我们可以计算的最大总和值是多少. 举例: 有一个数列:2 0 1 0 3 4 5 6,针对这个…

今天我们来聊一下数组的常用方法:split 返回值:一个新数组. 1.该方法可以直接调用不传任何值,则会直接将字符串转化成数组. var str = 'I love Javascript'; console.log(str.split()); //["I","love","Javascript"] 2.如果以不存在的字符串(或者正则匹配不到的字符串)进行分割,也会直接将字符串转成数组. 2.该方法有两个参数,第一个参数为必须参数(separato…

1.Array.from(objec,回调函数)将一个ArrayLike对象或者Iterable对象(类数组对象)转换成一个数组 1)该类数组对象必须具有length属性,用于指定数组的长度.如果没有length属性,那么转换后的数组是一个空数组 2)该类数组对象的属性名必须为数值型或字符串型的数字 //新的方法 //Array.from可以把一个类数组的对象转成数组 //该类数组对象的属性名必须为数值型或字符串型的数字 const obj = { 0:1, 1:22, 2:false, len…

2021.12.10 P2516 [HAOI2010]最长公共子序列(动态规划+滚动数组) https://www.luogu.com.cn/problem/P2516 题意: 给定字符串 \(S\) . \(T\) ,都以 \(.\) 结尾,求 \(S\) . \(T\) 最长公共子序列的长度及个数. 分析: 一顿操作猛如虎,一看分数250--爆零了.原本就没准备拿几分,结果令人心塞. 第一问就是求最长公共子序列长度,数据范围比较小, \(O(n^2)\) 就行,上来就是一顿树状数组+LIS,…