题目描述
数字是不断进入数组的，在每次添加一个新的数进入数组的同时返回当前新数组的中位数。
思路
用一个大顶堆一个小顶堆即可。
参考：滑动窗口的中位数
代码
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| class Help implements Comparable<Help>{
public int val;
public Help(int val) {
this.val = val;
}
@Override
public int compareTo(Help o) {
// TODO Auto-generated method stub
return this.val < o.val ? 1 :
this.val > o.val ? -1 : 0;
}
}
public int[] medianII(int[] nums) {
// init
int[] ret = new int[nums.length];
PriorityQueue<Integer> minHeap = new PriorityQueue<>();
PriorityQueue<Help> maxHeap = new PriorityQueue<>();
ret[0] = nums[0];
maxHeap.add(new Help(nums[0]));
// find mid after first
for(int i = 1; i < nums.length; i++) {
if(nums[i] > maxHeap.peek().val) {
minHeap.add(nums[i]);
}else {
maxHeap.add(new Help(nums[i]));
}
if(i % 2 == 0) {
if(maxHeap.size() > minHeap.size()) {
while((maxHeap.size() - minHeap.size()) != 1) {
minHeap.add(maxHeap.poll().val);
}
ret[i] = maxHeap.peek().val;
}else {
while((maxHeap.size() - minHeap.size()) != 1) {
maxHeap.add(new Help(minHeap.poll()));
}
ret[i] = maxHeap.peek().val;
}
}else {
if(maxHeap.size() > minHeap.size()) {
while(maxHeap.size() != minHeap.size()) {
minHeap.add(maxHeap.poll().val);
}
ret[i] = maxHeap.peek().val;
}else {
while(maxHeap.size() != minHeap.size()) {
maxHeap.add(new Help(minHeap.poll()));
}
ret[i] = maxHeap.peek().val;
}
}
}
return ret;
}
|
考差点