452. Minimum Number of Arrows to Burst Balloons

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Problem

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.

Example 1:

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).

Example 2:

Input: points = [[1,2],[3,4],[5,6],[7,8]]
Output: 4

Example 3:

Input: points = [[1,2],[2,3],[3,4],[4,5]]
Output: 2

Code

class Solution {
    public int findMinArrowShots(int[][] points) {
        Arrays.sort(points, (a, b) -> {
            // 不能使用 a[0] - b[0]
            // 数字太大会造成overflow
            // [[-2147483646,-2147483645],[2147483646,2147483647]]
            if (a[0] == b[0]) return 0;
            if (a[0] < b[0]) return -1;
            return 1;
        });
        
        int res = 1;
        int[] start = points[0];

        for(int i = 1; i < points.length; i++) {
            int[] curr = points[i];

            if(curr[0] > start[1]) {
                res++;
                start = curr;
            } else {
                start[0] = Math.max(start[0], curr[0]);
                start[1] = Math.min(start[1], curr[1]);
            }
        }
        
        return res;
    }
}