滑动窗口的中位数

给定一个包含 n 个整数的数组，和一个大小为 k 的滑动窗口,从左到右在数组中滑动这个窗口，找到数组中每个窗口内的中位数。(如果数组个数是偶数，则在该窗口排序数字后，返回第 N/2 个数字。)

您在真实的面试中是否遇到过这个题？ 

Yes

样例

对于数组 [1,2,7,8,5], 滑动大小 k = 3 的窗口时，返回 [2,7,7]

最初，窗口的数组是这样的：

[ | 1,2,7 | ,8,5] , 返回中位数 2;

接着，窗口继续向前滑动一次。

[1, | 2,7,8 | ,5], 返回中位数 7;

接着，窗口继续向前滑动一次。

[1,2, | 7,8,5 | ], 返回中位数 7;

import java.util.*;/** * Created by wb-xzc291800 on 2017/6/30. */public class Solution {        int mid = 0;        /**     * @param nums: A list of integers.     * @return: The median of the element inside the window at each moving.     */    public ArrayList<Integer> medianSlidingWindow(int[] nums, int k) {        mid = getMid(k);        ArrayList<Integer> a = new ArrayList<Integer>();        if (k == 0 || nums.length < mid) {            return a;        } else if (nums.length == mid || (nums.length > mid && nums.length < k)) {            a.add(nums[mid - 1]);            return a;        } else if (1 == k) {            for (int i : nums) {                a.add(i);            }            return a;        } else if (2 == k) {            for (int i = 0; i < nums.length - 1; i++) {                if (nums[i] > nums[i + 1]) {                    a.add(nums[i + 1]);                } else {                    a.add(nums[i]);                }            }            return a;        }                LinkedList<Integer> sub = subArray(nums, k);                a.add(sub.get(mid - 1));        for (int i = k; i < nums.length; i++) {            start = 0;            end = 0;            insert(sub, nums[i]);            sub.remove(sub.indexOf(nums[i - k]));            a.add(sub.get(mid - 1));        }        return a;    }        int getMid(int k) {        int mid = 0;        if (k % 2 == 1) {            mid = (k + 1) / 2;        } else {            mid = k / 2;        }        return mid;    }    /**     * 截取首个视窗数组,并且排序     * @param nums     * @param k     * @return     */    LinkedList<Integer> subArray(int[] nums, int k) {        LinkedList<Integer> ret = new LinkedList<Integer>();        for (int i = 0; i < k; i++) {            ret.add(nums[i]);        }        Collections.sort(ret);        return ret;    }    int start = 0;    int end = 0;    /**     * 二分法查找插入值     * @param linkedList     * @param val     */    void insert(LinkedList<Integer> linkedList, int val) {        if (end == 0) {            end = linkedList.size() - 1;        }        if (end - start <= 1) {            if (linkedList.get(start) >= val) {                linkedList.add(start, val);                return;            }            if (linkedList.get(end) <= val || end == linkedList.size()) {                linkedList.addLast(val);                return;            }            linkedList.add(end,val);            return;        }        int mid = (start + end) / 2;        int x = linkedList.get(mid);        if (x > val) {            end = mid;        } else {            start = mid;        }        insert(linkedList, val);    }}