LeetCode452, use the least number of arrows to detonate the balloon (greedy selection interval scheduling)

Title description

https://leetcode-cn.com/problems/minimum-number-of-arrows-to-burst-balloons/

solution

Same as the essence of the previous question.

class Solution {
    
    
    public int findMinArrowShots(int[][] points) {
    
    
        //贪心选择，尽可能地一次射爆交叠最多的气球的坐标范围
        //剩余不交叠的区间，一支射爆一个

        //从可视化图上看，我们只需要求出最大不交叠的组合即可
        if(points==null || points.length==0) return 0;
        //1、对活动的end时间进行升序排序
        Arrays.sort(points,new Comparator<int[]>(){
    
    
            public int compare(int[]a,int[]b){
    
    
                return a[1]>b[1]?1:-1;//升序排序
            }
        });
        //贪心选择
        int count = 1;//必须
        int end = points[0][1];
        for(int i=1;i<points.length;i++){
    
    
            int start = points[i][0];
            if(end<start){
    
    //相等也是交叠
                end = points[i][1];//更新
                count++;
            }
        }
        return count;

    }
}

After reading the following questions and comments, a very interesting but serious question: When

sorting, don't write it like the following to save trouble. The data will overflow. Or honestly compare the size to return positive and negative.

 Arrays.sort(points,new Comparator<int[]>(){
    
    
            public int compare(int[]a,int[]b){
    
    
                //return a[1]>b[1]?1:-1;//升序排序
                return a[1]-b[1];
            }
        });