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…
[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…
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…
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…
Given an array A of non-negative integers, half of the integers in A are odd, and half of the integers are even. Sort the array so that whenever A[i] is odd, i is odd; and whenever A[i] is even, i is even. You may return any answer array that satisfi…
题目描述: 我的分析:题目要求将奇数放在偶数的前面,没有要求将奇数或偶数排序,因此我可以设置两个指针,一个(i)指向数组第一个数字,另一个(j)指向数组的最后一个数字,因为奇数要放在前面,所以从后往前找奇数,从前往后找偶数,找到后将这两个数字进行交换,直到i == j. 我的代码: public class Solution { /* * @param nums: an array of integers * @return: nothing */ public void partitionAr…
A. Array with Odd Sum time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output You are given an array…
