问题描述

There are some spherical balloons spread in two-dimensional space. For each balloon, provided input is the start and end coordinates of the horizontal diameter. Since it’s horizontal, y-coordinates don’t matter, and hence the x-coordinates of start and end of the diameter suffice. The start is always smaller than the end.

An arrow can be shot up exactly vertically from different points along the x-axis. A balloon with xstart and xend bursts by an arrow shot at x if xstart ≤ x ≤ xend. There is no limit to the number of arrows that can be shot. An arrow once shot keeps traveling up infinitely.

Given an array points where points[i] = [xstart, xend], return the minimum number of arrows that must be shot to burst all balloons.

测试样例

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Input: points = [[10,16],[2,8],[1,6],[7,12]]
Output: 2
Explanation: One way is to shoot one arrow for example at x = 6 (bursting the balloons [2,8] and [1,6]) and another arrow at x = 11 (bursting the other two balloons).
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Input: points = [[1,2],[3,4],[5,6],[7,8]]
Output: 4
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Input: points = [[1,2],[2,3],[3,4],[4,5]]
Output: 2

说明

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1 <= points.length <= 10^4
points[i].length == 2
-2^31 <= xstart < xend <= 2^31 - 1

解题

思路

区间重叠情况，一般可以使用贪心算法。

把所有的区间按照右边界进行排序。遍历时只要向当前气球的右边界射箭，就能打破尽可能多的气球。然后依次移动射箭的位置，进行统计即可。

补充：

  1. 先排序
  2. 贪心算法
  3. 时间复杂度 O(nlogn)（排序）

代码

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import java.util.Arrays;

class Solution {
    public int findMinArrowShots(int[][] points) {
        
        // 按区间右侧数字，从小到大排序
        Arrays.sort(points, (arr1, arr2) -> arr1[1] - arr2[1]);
        
        // arrow 数量
        int cnt_arrow = 1;
        // arrow 位置
        int p_arrow = points[0][1];
        // 遍历气球
        for(int i=0; i<points.length; ++i) {
            // 第 i 个气球的左右边界
            int p_left = points[i][0], p_right = points[i][1];
            
            // arrow 坐标在当前气球边界外
            if(p_arrow < p_left || p_right < p_arrow) {
                p_arrow = p_right;  // 将 arrow 坐标设为当前气球右侧边界
                ++cnt_arrow;        // arrow 数量增加
            } // arrow 坐标在当前气球边界内，直接下一个
            
        }
        
        return cnt_arrow;
    }
}