LintCode 373: Partition Array 题目描述 分割一个整数数组,使得奇数在前偶数在后. 样例 给定[1, 2, 3, 4],返回[1, 3, 2, 4]. Thu Feb 23 2017 思路 简单题,可以很自然地想到再用一个答案数组,从头到尾遍历一遍,遇到奇数就放到答案数组的前面,遇到偶数就放到答案数组的后面. 还有另一种方法,跟快速排序的形式有点像,即从前面找到一个偶数,同时从后面找到一个奇数,将两个数调换. 虽然两种方法的时间复杂度都是\(O(n)\),但是第二种方…

Description Partition an integers array into odd number first and even number second. Example Given [1, 2, 3, 4], return [1, 3, 2, 4] Challenge Do it in-place. 解题:将一个数组中的奇数和偶数分开,原地分开,不申请额外的空间.思路是,记录好最后一个奇数的后一个位置,也可以理解为偶数中的第一个·.遍历数组,如果当前数组中的元素是偶数,那么什么…

Given an array "nums" of integers and an int "k", Partition the array (i.e move the elements in "nums") such that, * All elements < k are moved to the left * All elements >= k are moved to the right Return the partition…

题目 数组划分 给出一个整数数组nums和一个整数k.划分数组(即移动数组nums中的元素),使得: 所有小于k的元素移到左边 所有大于等于k的元素移到右边 返回数组划分的位置,即数组中第一个位置i,满足nums[i]大于等于k. 您在真实的面试中是否遇到过这个题? Yes 样例 给出数组nums=[3,2,2,1]和 k=2,返回 1 注意 你应该真正的划分数组nums,而不仅仅只是计算比k小的整数数,如果数组nums中的所有元素都比k小,则返回nums.length. 挑战 要求在原地使用O…

题目: 奇偶分割数组 分割一个整数数组,使得奇数在前偶数在后. 样例 给定 [1, 2, 3, 4],返回 [1, 3, 2, 4]. 挑战 在原数组中完成,不使用额外空间. 解题: 一次快速排序就可以得到结果 Java程序: public class Solution { /** * @param nums: an array of integers * @return: nothing */ public void partitionArray(int[] nums) { // write…

One pass in-place solution: all swaps. class Solution { public: /** * @param nums: a vector of integers * @return: nothing */ void partitionArray(vector<int> &nums) { size_t len = nums.size(); , io = , ie = len - ; while (io < ie) { int v = n…

[题目描述] Partition an integers array into odd number first and even number second. 分割一个整数数组,使得奇数在前偶数在后. [题目链接] www.lintcode.com/en/problem/partition-array-by-odd-and-even/ [题目解析] 1.将数组中的奇数和偶数分开,使用『两根指针』的方法,用快排的思路,右指针分别从数组首尾走起 2.左指针不断往右走直到遇到偶数,右指针不断往左走直…

Given an array nums of integers and an int k, partition the array (i.e move the elements in "nums") such that: All elements < k are moved to the left All elements >= k are moved to the right Return the partitioning index, i.e the first ind…

Given an array A, partition it into two (contiguous) subarrays left and right so that: Every element in left is less than or equal to every element in right. left and right are non-empty. left has the smallest possible size. Return the length of left…

Given an array A of integers, return true if and only if we can partition the array into three non-emptyparts with equal sums. Formally, we can partition the array if we can find indexes i+1 < j with (A[0] + A[1] + ... + A[i] == A[i+1] + A[i+2] + ...…

题目要求 Given an array A of integers, return true if and only if we can partition the array into three non-empty parts with equal sums. Formally, we can partition the array if we can find indexes i+1 < j with (A[0] + A[1] + ... + A[i] == A[i+1] + A[i+2]…

Given an array A, partition it into two (contiguous) subarrays left and right so that: Every element in left is less than or equal to every element in right. left and right are non-empty. left has the smallest possible size. Return the length of left…

Given an array A of integers, return true if and only if we can partition the array into three non-empty parts with equal sums. Formally, we can partition the array if we can find indexes i+1 < j with (A[0] + A[1] + ... + A[i] == A[i+1] + A[i+2] + ..…

2组: ［抄题］: 给出一个整数数组 nums 和一个整数 k.划分数组(即移动数组 nums 中的元素),使得: 所有小于k的元素移到左边 所有大于等于k的元素移到右边 返回数组划分的位置,即数组中第一个位置 i,满足 nums[i] 大于等于 k. ［思维问题］: 想不到两个小人的partition ［一句话思路］: 两个小人的partition,不用排序 ［输入量］:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入): ［画图］: ［一刷］: 直接调换就行了…

Partition an integers array into odd number first and even number second. Example Given [, , , ], , , , ] Challenge Do it in-place. 将数组中的奇数和偶数分开,使用『两根指针』的方法最为自然,奇数在前,偶数在后,若不然则交换之. JAVA: public class Solution { /** * @param nums: an array of integers…

Given an array A, partition it into two (contiguous) subarrays left and right so that: Every element in left is less than or equal to every element in right. left and right are non-empty. left has the smallest possible size. Return the length of left…

原题链接在这里:https://leetcode.com/problems/partition-array-for-maximum-sum/ 题目: Given an integer array A, you partition the array into (contiguous) subarrays of length at most K.  After partitioning, each subarray has their values changed to become the ma…

题目如下: Given an integer array A, you partition the array into (contiguous) subarrays of length at most K.  After partitioning, each subarray has their values changed to become the maximum value of that subarray. Return the largest sum of the given arr…

题目如下: Given an array A of integers, return true if and only if we can partition the array into three non-emptyparts with equal sums. Formally, we can partition the array if we can find indexes i+1 < j with (A[0] + A[1] + ... + A[i] == A[i+1] + A[i+2]…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.com/problems/partition-array-into-three-parts-with-equal-sum/ 题目描述 Given an array A of integers, return true if and only if we can partition the array…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址: https://leetcode.com/problems/partition-array-into-disjoint-intervals/description/ 题目描述: Given an array A, partition it into two (contiguous) subarrays left and right so that: Every elemen…

Given an integers array A. Define B[i] = A[0] * ... * A[i-1] * A[i+1] * ... * A[n-1], calculate B WITHOUT divide operation. Have you met this question in a real interview? Yes Example For A = [1, 2, 3], return [6, 3, 2]. LeetCode上的原题,请参见我之前的博客Product…

两个排好序的数组, A有m个数, 长度够长, B有n个数, 长度为n, 将B放到A里, 不用buffer数组(临时数组), 就用A将两个合在一起, 同时排好序. 方法: 从右边将两个数组里最大的数取出放入A的m+n-1的index处, 从右往左放, 两指针i和j, 直到i和j都等于-1. 注意:时间和空间复杂度: O(m+n) 和 O(1). 空间复杂度即为额外用到的, 这里为0, 即O(1). class Solution { /** * @param A: sorted integer ar…

Given an integers array A. Define B[i] = A[0] * ... * A[i-1] * A[i+1] * ... * A[n-1], calculate B without divide operation. Example For A=[1, 2, 3], B is [6, 3, 2] 非常典型的Forward-Backward Traversal 方法: 但是第一次做的时候还是忽略了一些问题:比如A.size()==1时,答案应该是空[] public…

Merge two given sorted integer array A and B into a new sorted integer array. Example A=[1,2,3,4] B=[2,4,5,6] return [1,2,2,3,4,4,5,6] Challenge How can you optimize your algorithm if one array is very large and the other is very small? 没什么好说的,Arrayl…

题目: 链表划分 给定一个单链表和数值x,划分链表使得所有小于x的节点排在大于等于x的节点之前. 你应该保留两部分内链表节点原有的相对顺序. 样例 给定链表 1->4->3->2->5->2->null,并且 x=3 返回 1->2->2->4->3->5->null 解题: 上面返回的结果好多不对的应该是: 1->2->2->3->4->5->null 想了好久不知道怎么搞,九章看到的程序,竟然搞…

C++ /** * Definition of TreeNode: * class TreeNode { * public: * int val; * TreeNode *left, *right; * TreeNode(int val) { * this->val = val; * this->left = this->right = NULL; * } * } */ class Solution { public: /** * @param A: A sorted (increasi…

简单题,暴力找出来就行. class Solution: def canThreePartsEqualSum(self, A: List[int]) -> bool: s = sum(A) if s % 3 > 0: return False s /= 3 size = len(A) t = 0 a, b = -1, -1 for i in range(size): t += A[i] if t == s: a = i break t = 0 for j in range(size - 1,…

题目如下: 解题思路:题目要求的是在数组中找到一个下标最小的index,使得index左边(包括自己)子序列的最大值小于或者等于右边序列的最小值.那么我们可以先把数组从最左边开始到数组最右边所有子序列的最大值都求出来存入leftMax数组,同时也把从数组最右边开始到最左边的所有子序列的最小值求出来存入rightMin数组,最后找出最小的i使得leftMax[i] <= rightMin[i+1]即可. 代码如下: class Solution(object): def partitionDisj…

2020-02-10 22:16:50 问题描述: 问题求解: 解法一:MultiSet O(nlog) 看了下数据规模,第一个想到的是multiset,肯定可以ac的,就直接敲了出来. public int partitionDisjoint(int[] A) { TreeMap<Integer, Integer> map = new TreeMap<>(); for (int num : A) map.put(num, map.getOrDefault(num, 0) + 1)…