1. min/max heap
看到K神马的基本上就是min/max heap.
（1） Find the K closest points to the origin in a 2D plane, given an array containing N points.
1 public static List<Point> findKClosest(Point[] p, int k) {
2
3 List<Point> res = new ArrayList<Point>();
4 Comparator<Point> comparator = new Comparator<Point>(){
5 //@Override
6 public int compare(Point a, Point b) {
7 return (int) ( a.x*a.x + a.y*a.y- b.x*b.x -b.y*b.y);
8 }
9 };
10
11 PriorityQueue<Point> minheap = new PriorityQueue<Point>(comparator);
12 for (Point temp : p) {
13 minheap.add(temp);
14 }
15 for (int i = 0; i < k; i++) {
16 res.add(minheap.poll());
17 }
18
19 return res;
20 }
version1
如果只能用k个位置保留在有线队列里面的话，那么前k个可以直接加入，后面k个的话，如果比最后一个小，那么把最后1个移除，然后放入当前的点.
1 public List<Point> findKClosest(Point[] p, int k) {
2 PriorityQueue<Point> pq = new PriorityQueue<>(10, new Comparator<Point>() {
3 @Override
4 public int compare(Point a, Point b) {
5 return (b.x * b.x + b.y * b.y) - (a.x * a.x + a.y * a.y);
6 }
7 });
8
9 for (int i = 0; i < p.length; i++) {
10 if (i < k)
11 pq.offer(p[i]);
12 else {
13 Point tmp = pq.peek();
14 if ((p[i].x * p[i].x + p[i].y * p[i].y) - (tmp.x * tmp.x + tmp.y * tmp.y) < 0) {
15 pq.poll();
16 pq.offer(p[i]);
17 }
18 }
19 }
20
21 List<Point> x = new ArrayList<>();
22 while (!pq.isEmpty())
23 x.add(pq.poll());
24
25 return x;
26 }
version2
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2. 队列 queue
支持操作 o(1) push, o(1) pop, o(1) top
BFS的主要数据结构 多做做bfs题就好了
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3. 栈 stack
支持操作 o(1) push, o(1) pop, o(1) top
非递归实现dfs的重要数据结构
(1) min stack
1 public class MinStack {
2
3 /** initialize your data structure here. */
4 static Stack<Integer> numStack;
5 static Stack<Integer> minStack;
6
7 public MinStack() {
8 numStack = new Stack<Integer>();
9 minStack = new Stack<Integer>();
10 }
11
12 public void push(int x) {
13 numStack.push(x);
14 if (minStack.empty() || x <= minStack.peek()) {
15 minStack.push(x);
16 }
17 }
18
19 public void pop() {
20 if (numStack.peek().equals(minStack.peek())) {
21 //note!!! because Stack contains integer which is an object, is use == will test its pointer!!!
22 minStack.pop();
23 numStack.pop();
24
25 } else {
26 numStack.pop();
27 }
28 }
29
30 public int top() {
31 return numStack.peek();
32 }
33
34 public int getMin() {
35 return minStack.peek();
36 }
37 }
min stack
(2) implement queue by two stack
1 class MyQueue {
2 private Stack<Integer> stack1 = new Stack<Integer>();
3 private Stack<Integer> stack2 = new Stack<Integer>();
4
5 // Push element x to the back of queue.
6 public void push(int x) {
7 if (!stack2.empty()){
8 move();
9 }
10 stack1.push(x);
11
12 }
13 private void move() {
14 if (!stack2.empty()) {
15 while (!stack2.empty()) {
16 stack1.push(stack2.pop());
17 }
18 return;
19 } else {
20 while (!stack1.empty()) {
21 stack2.push(stack1.pop());
22 }
23 return;
24 }
25 }
26
27 // Removes the element from in front of queue.
28 public void pop() {
29 if (!stack1.empty()) {
30 move();
31 }
32 stack2.pop();
33 }
34
35 // Get the front element.
36 public int peek() {
37 if (!stack1.empty()) {
38 move();
39 }
40 return stack2.peek();
41 }
42
43 // Return whether the queue is empty.
44 public boolean empty() {
45 if (!stack1.empty() || !stack2.empty()) {
46 return false;
47 }
48 return true;
49 }
50 }
(3)Largest rectangle in histogram
Given n non-negative integers representing the histogram‘s bar height where the width of each bar is 1, find the area of largest rectangle in the histogram.
Above is a histogram where width of each bar is 1, given height = [2,1,5,6,2,3].
The largest rectangle is shown in the shaded area, which has area = 10 unit.
For example,
Given heights = [2,1,5,6,2,3],
return 10.
暴力方法：
for矩阵的起点
需要用到“单调栈”
这道题的核心点是：最矮的木头！
1. //for矩阵的最爱的那根木头，也就是矩阵的高度
2.记住，单调栈能够做到找到左边比一个比当前小的点，能找到右边第一个比它小的点
3. 单调栈里实际上存的是下标，在比较增序关系的时候才用值去比较
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