Lintcode: Partition Array
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 index "i" nums[i] >= k.
Note
You should do really partition in array "nums" instead of just counting the numbers of integers smaller than k.
If all elements in "nums" are smaller than k, then return "nums.length"
Example
If nums=[3,2,2,1] and k=2, a valid answer is 1.
Challenge
Can you partition the array in-place and in O(n)?
Quick Sort 一样的做法,只是有两种情况特殊处理:我第一次做的时候没有考虑到
1. all elements in nums are greater than or equal to k, l pointer never shift, should return l
2. all elements in nums are smaller than k, r pointer never shift, shoud return r+1
第一次做法(稍次)
public class Solution {
/**
*@param nums: The integer array you should partition
*@param k: As description
*return: The index after partition
*/
public int partitionArray(ArrayList<Integer> nums, int k) {
//write your code here
if (nums==null || nums.size()==0) return 0;
int l=0, r=nums.size()-1;
while (true) {
while (l<r && nums.get(r)>=k) {
r--;
}
while (l<r && nums.get(l)<k) {
l++;
}
if (l == r) break;
swap(l, r, nums);
}
if (l==0 && nums.get(l)>=k) return r;
if (r==nums.size()-1 && nums.get(l)<k) return r+1;
return r+1;
}
public void swap(int l, int r, ArrayList<Integer> nums) {
int temp = nums.get(l);
nums.set(l, nums.get(r).intValue());
nums.set(r, temp);
}
}
第二次做法(推荐): 只要l,r 都动过,l停的位置就是first index that nums[i] >= k, 一般情况return l就好了
单独讨论l或者r没有动过的情况,l没有动过的情况还是return l, r没有动过的情况return r+1
public class Solution {
/**
*@param nums: The integer array you should partition
*@param k: As description
*return: The index after partition
*/
public int partitionArray(int[] nums, int k) {
//write your code here
if (nums==null || nums.length==0) return 0;
int l=0, r=nums.length-1;
while (true) {
while (l<r && nums[l]<k) {
l++;
}
while (l<r && nums[r]>=k) {
r--;
}
if (l == r) break;
swap(l, r, nums);
}
//if (l==0 && nums[l]>=k) return l;
if (r==nums.length-1 && nums[r]<k) return r+1;
return l;
}
public void swap(int l, int r, int[] nums) {
int temp = nums[l];
nums[l] = nums[r];
nums[r] = temp;
}
}
Lintcode: Partition Array的更多相关文章
- LintCode "Partition Array by Odd and Even"
One pass in-place solution: all swaps. class Solution { public: /** * @param nums: a vector of integ ...
- LintCode 373: Partition Array
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 ...
- lintcode 中等题:partition array 数组划分
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, ...
- 373. Partition Array by Odd and Even【LintCode java】
Description Partition an integers array into odd number first and even number second. Example Given ...
- lintcode 容易题:Partition Array by Odd and Even 奇偶分割数组
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 ...
- Lintcode373 Partition Array by Odd and Even solution 题解
[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 ...
- Partition Array
Given an array nums of integers and an int k, partition the array (i.e move the elements in "nu ...
- [Swift]LeetCode915.将分区数组分成不相交的间隔 | Partition Array into Disjoint Intervals
Given an array A, partition it into two (contiguous) subarrays left and right so that: Every element ...
- [Swift]LeetCode1013. 将数组分成和相等的三个部分 | Partition Array Into Three Parts With Equal Sum
Given an array A of integers, return true if and only if we can partition the array into three non-e ...
随机推荐
- 二 mybatis 动态sql
动态sql应用 一 .什么是动态sql 1.where条件 动态查询 根据姓名或年龄或地址查询 UserMapper.xml 1 <select id="findUser" ...
- memache session
Memcache和PHP memcach扩展安装请见http://koda.iteye.com/blog/665761 设置session用memcache来存储 方法I: 在 php.ini 中全局 ...
- libopencv_highgui.a(window_gtk.cpp.o): undefined reference to symbol 'g_type_check_instance_is_a'
libopencv_highgui.a(window_gtk.cpp.o): undefined reference to symbol 'g_type_check_instance_is_a' 尝试 ...
- [收藏]Asp.net MVC生命周期
一个HTTP请求从IIS移交到Asp.net运行时,Asp.net MVC是在什么时机获得了控制权并对请求进行处理呢?处理过程又是怎样的? 以IIS7中asp.net应用程序生命周期为例,下图是来自M ...
- ImageX用来做Windows OEM部署
https://technet.microsoft.com/en-us/library/cc722145(v=ws.10).aspx http://download.csdn.net/user/phc ...
- js严格模式“use strict”
js的严格模式会放弃js中的一些不正规的写法,参考 http://www.cnblogs.com/God-Shell/p/3139329.html: 使用声明"use strict&quo ...
- 过滤android应用列表(区分系统应用、第三方应用、sd卡中的应用)
if ((app.flags & ApplicationInfo.FLAG_SYSTEM) != 0) { // 系统程序 }else if ((app.flags & Applica ...
- iOS开发教程之:iPhone开发环境搭建
安装条件: 硬件:一台拥有支持虚拟技术的64位双核处理器和2GB以上内存的PC. 注意:运行MAC OS,需要电脑支持虚拟技术(VT),安装时,需要将VT启动,在BIOS中开启. 软件: Window ...
- 采用asyncore进行实时同步
最近在维护项目的时候,发现某个实时数据同步功能非常容易失败,故静下心来彻底弄清楚该设计的实现原理,以及其中用到的python异步sockethandler : asyncore. 实时数据同步功能的设 ...
- [LeetCode]题解(python):111 Minimum Depth of Binary Tree
题目来源 https://leetcode.com/problems/minimum-depth-of-binary-tree/ Given a binary tree, find its minim ...