区间问题

区间问题之前自己写的时候都做出来了，但是一到oa的时候就懵，已经以为这种问题挂了两次了。现在就先总结一下

1. 452. Minimum Number of Arrows to Burst Balloons

There are some spherical balloons taped onto a flat wall that represents the XY-plane. The balloons are represented as a 2D integer array points where points[i] = [xstart, xend] denotes a balloon whose horizontal diameter stretches between xstart and xend. You do not know the exact y-coordinates of the balloons.

Arrows can be shot up directly vertically (in the positive y-direction) from different points along the x-axis. A balloon with xstart and xend is burst by an arrow shot at x if xstart <= x <= xend. There is no limit to the number of arrows that can be shot. A shot arrow keeps traveling up infinitely, bursting any balloons in its path.

Given the array points, return the minimum number of arrows that must be shot to burst all balloons.

 

Example 1:

Input: points = [[10,16],[2,8],[1,6],[7,12]]
Output: 2
Explanation: The balloons can be burst by 2 arrows:
- Shoot an arrow at x = 6, bursting the balloons [2,8] and [1,6].
- Shoot an arrow at x = 11, bursting the balloons [10,16] and [7,12].

Example 2:

Input: points = [[1,2],[3,4],[5,6],[7,8]]
Output: 4
Explanation: One arrow needs to be shot for each balloon for a total of 4 arrows.

Example 3:

Input: points = [[1,2],[2,3],[3,4],[4,5]]
Output: 2
Explanation: The balloons can be burst by 2 arrows:
- Shoot an arrow at x = 2, bursting the balloons [1,2] and [2,3].
- Shoot an arrow at x = 4, bursting the balloons [3,4] and [4,5].

两种方法：

1. 按照区间起点从小到大排序
2. 按照区间终点从小到大排序

这道题其实算是合并区间取交集 按照起点排序，因为区间起点是依次增加的，end就取最小值。下一次起点大于end，res就++，并且重制end

class Solution {
    public int findMinArrowShots(int[][] points) {
        Arrays.sort(points, (a, b) -> Integer.compare(a[0], b[0]));
        int res = 1;
        long end = points[0][1];
        for(int i = 1; i < points.length; i++) {
            if(points[i][0] > end) {
                res++;
                end = points[i][1];
            } else {
                end = Math.min(end, points[i][1]);
            }
        }

        return res;
    }
}

而如果按照区间终点排序： 因为start是一直小于end的，所以只要 下一个区间起点的值大于end 那么res++，然后重置end

class Solution {
    public int findMinArrowShots(int[][] points) {
        Arrays.sort(points, (a, b) -> Integer.compare(a[1], b[1]));
        
        int res = 1;
        int end = points[0][1];

        for(int i = 1; i < points.length; i++) {
            if(points[i][0] > end) {
                res++;
                end = points[i][1];
            }
        }

        return res;
    }
}

Meeting rooms 2

Given an array of meeting time intervals intervals where intervals[i] = [starti, endi], return the minimum number of conference rooms required.

 

Example 1:

Input: intervals = [[0,30],[5,10],[15,20]]
Output: 2

Example 2:

Input: intervals = [[7,10],[2,4]]
Output: 1

class Solution {
    public int minMeetingRooms(int[][] intervals) {
        Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
        PriorityQueue<Integer> pq = new PriorityQueue<>();
        int maxCount = 0;
        for(int[] interval: intervals) {
            if(!pq.isEmpty() && pq.peek() <= interval[0]) {
                pq.poll();
            }
            pq.add(interval[1]);
        }
        return pq.size();
    }
}