Numbers keep coming, return the median of numbers at every time a new number added.

Example

For numbers coming list: [1, 2, 3, 4, 5], return [1, 1, 2, 2, 3]

For numbers coming list: [4, 5, 1, 3, 2, 6, 0], return [4, 4, 4, 3, 3, 3, 3]

For numbers coming list: [2, 20, 100], return [2, 2, 20]

Challenge

O(nlogn) time

Clarification

What's the definition of Median?

  • Median is the number that in the middle of a sorted array. If there are n numbers in a sorted array A, the median is A[(n-1)/2].
  • For example, if A=[1,2,3], median is 2. If A=[1,19], median is 1

Analysis:

We maintain a max heap and a min heap. At any index i, the max heap stores the elements small or equal than the current median and the min heap stores the elements that are larger than the current median. Because in the problem, we select the median as at the position (n-1)/2, so we should always keep the size of max heap equal or one larger than the min heap. At odd index, we try to rebalance the size of max heap and min heap, while at even index, we try to make the size of max heap one larger than that of the min heap. By doing this, at each position, after inserting A[i] into the heaps, the root value of the max heap is the median.

NOTE: if we select (n-1)/2+1 as median, we can then keep the min heap equal or one larger than the max heap, the root value of the min heap is the median.

Solution:

I implement a Heap class with only insert and popHeapRoot functions.

 class Heap{
private int[] nodes;
private int size;
private boolean isMaxHeap; public Heap(int capa, boolean isMax){
nodes = new int[capa];
size = 0;
isMaxHeap = isMax;
} public boolean isEmpty(){
if (size==0) return true;
else return false;
} public int getHeapRootValue(){
//should throw exception when size==0;
return nodes[0];
} private void swap(int x, int y){
int temp = nodes[x];
nodes[x] = nodes[y];
nodes[y] = temp;
} public boolean insert(int val){
if (size==nodes.length) return false;
size++;
nodes[size-1]=val;
//check its father iteratively.
int cur = size-1;
int father = (cur-1)/2;
while (father>=0 && ((isMaxHeap && nodes[cur]>nodes[father]) || (!isMaxHeap && nodes[cur]<nodes[father]))){
swap(cur,father);
cur = father;
father = (cur-1)/2;
}
return true;
} private void shiftDown(int ind){
int left = (ind+1)*2-1;
int right = (ind+1)*2;
while (left<size || right<size){
if (isMaxHeap){
int leftVal = (left<size) ? nodes[left] : Integer.MIN_VALUE;
int rightVal = (right<size) ? nodes[right] : Integer.MIN_VALUE;
int next = (leftVal>=rightVal) ? left : right;
if (nodes[ind]>nodes[next]) break;
else {
swap(ind,next);
ind = next;
left = (ind+1)*2-1;
right = (ind+1)*2;
}
} else {
int leftVal = (left<size) ? nodes[left] : Integer.MAX_VALUE;
int rightVal = (right<size) ? nodes[right] : Integer.MAX_VALUE;
int next = (leftVal<=rightVal) ? left : right;
if (nodes[ind]<nodes[next]) break;
else {
swap(ind,next);
ind = next;
left = (ind+1)*2-1;
right = (ind+1)*2;
}
}
}
} public int popHeapRoot(){
//should throw exception, when heap is empty. int rootVal = nodes[0];
swap(0,size-1);
size--;
if (size>0) shiftDown(0);
return rootVal;
}
} public class Solution {
/**
* @param nums: A list of integers.
* @return: the median of numbers
*/
public int[] medianII(int[] nums) {
int[] res = new int[nums.length];
Heap maxHeap = new Heap(nums.length,true);
Heap minHeap = new Heap(nums.length,false);
maxHeap.insert(nums[0]);
res[0] = nums[0];
for (int i=1;i<nums.length;i++)
if (i %2 == 1) { //i is odd index.
int median = maxHeap.getHeapRootValue();
if (nums[i]<median){
maxHeap.popHeapRoot();
minHeap.insert(median);
maxHeap.insert(nums[i]);
res[i] = maxHeap.getHeapRootValue();
} else {
res[i] = median;
minHeap.insert(nums[i]);
}
} else { //i is even index.
int median = maxHeap.getHeapRootValue();
if (nums[i]<median){
maxHeap.insert(nums[i]);
} else {
minHeap.insert(nums[i]);
int val = minHeap.popHeapRoot();
maxHeap.insert(val);
}
res[i] = maxHeap.getHeapRootValue();
} return res;
}
}

LintCode-Median II的更多相关文章

  1. [LintCode] Median of Two Sorted Arrays 两个有序数组的中位数

    There are two sorted arrays A and B of size m and n respectively. Find the median of the two sorted ...

  2. Lintcode: Median

    Given a unsorted array with integers, find the median of it. A median is the middle number of the ar ...

  3. [LintCode] Permutations II

    Given a collection of numbers that might contain duplicates, return all possible unique permutations ...

  4. LintCode: Median of two Sorted Arrays

    求第k个值 1.归并排序 归并到第k个值为止 时间复杂度:O(k) class Solution { public: // merge-sort to find K-th value double h ...

  5. [算法专题] 深度优先搜索&回溯剪枝

    1. Palindrome Partitioning https://leetcode.com/problems/palindrome-partitioning/ Given a string s, ...

  6. lintcode 最长上升连续子序列 II(二维最长上升连续序列)

    题目链接:http://www.lintcode.com/zh-cn/problem/longest-increasing-continuous-subsequence-ii/ 最长上升连续子序列 I ...

  7. Lintcode: Sort Colors II 解题报告

    Sort Colors II 原题链接: http://lintcode.com/zh-cn/problem/sort-colors-ii/# Given an array of n objects ...

  8. Lintcode: Majority Number II 解题报告

    Majority Number II 原题链接: http://lintcode.com/en/problem/majority-number-ii/# Given an array of integ ...

  9. leetcode 293.Flip Game(lintcode 914) 、294.Flip Game II(lintcode 913)

    914. Flip Game https://www.cnblogs.com/grandyang/p/5224896.html 从前到后遍历,遇到连续两个'+',就将两个加号变成'-'组成新的字符串加 ...

  10. Lintcode 150.买卖股票的最佳时机 II

    ------------------------------------------------------------ 卧槽竟然连题意都没看懂,百度了才明白题目在说啥....我好方啊....o(╯□ ...

随机推荐

  1. 1369 xth 砍树

    1369 xth 砍树  时间限制: 1 s  空间限制: 128000 KB  题目等级 : 钻石 Diamond 题解       题目描述 Description 在一个凉爽的夏夜,xth 和 ...

  2. Table of Contents - ActiveMQ

    Getting Started ActiveMQ 的安装 Hello World Configuring Standard ActiveMQ Components Connecting to Acti ...

  3. Android聊天界面刷新消息

    今天,我想来分享一下自己初用线程的感受,虽然写法略显粗糙,并没有用线程Thread中核心的Looper,MessageQueue消息队列这些知识,正因为是初学线程,所以就只用最基础的来写了,慢慢学习优 ...

  4. php生成 优惠券 激活码

    /** * 生成vip激活码 * @param int $nums 生成多少个优惠码 * @param array $exist_array 排除指定数组中的优惠码 * @param int $cod ...

  5. ios警告:Category is implementing a method which will also be implemented by its primary class 引发的相关处理

    今天在处理项目中相关警告的时候发现了很多问题,包括各种第三方库中的警告,以及各种乱七八糟的问题  先说说标题中的问题  Category is implementing a method which ...

  6. ApplicationContext容器的设计原理

    1.在ApplicationContext容器中,我们以常用的FileSystemXmlApplicationContext的实现为例来说明ApplicationContext容器的设计原理. 2.在 ...

  7. android 数据库_sql语句总结

    表的创建db.execSQL("create table info(_id integer primary key autoincrement,name varchar(20)") ...

  8. Linux读写锁的使用

    读写锁是用来解决读者写者问题的,读操作可以共享,写操作是排它的,读可以有多个在读,写只有唯一个在写,写的时候不允许读. 具有强读者同步和强写者同步两种形式: 强读者同步:当写者没有进行写操作时,读者就 ...

  9. 暑假集训(4)第五弹——— 数论(hdu1222)

    题意概括:那天以后,你好说歹说,都快炼成三寸不烂之舍之际,小A总算不在摆着死人脸,鼓着死鱼眼.有了点恢复的征兆.可孟子这家伙说的话还是有点道理,那什么天将降....额,总之,由于贤者法阵未完成,而小A ...

  10. Android基本知识

         Android是Google公司于2007年发布的基于Linux内核的手机操作系统.应用层主要以java为编程语言,应用层分为两层,函数层(Library) 和虚拟机(Virtual).中间 ...