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…

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…

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…

[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 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] + ..…