452 Minimum Number of Arrows to Burst Balloons 用最少数量的箭引爆气球
Description:
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:
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
Example 4:
Input: points = [[1,2]]
Output: 1
Example 5:
Input: points = [[2,3],[2,3]]
Output: 1
Constraints:
0 <= points.length <= 10^4
points[i].length == 2
-2^31 <= xstart < xend <= 2^31 - 1
题目描述:
在二维空间中有许多球形的气球。对于每个气球，提供的输入是水平方向上，气球直径的开始和结束坐标。由于它是水平的，所以纵坐标并不重要，因此只要知道开始和结束的横坐标就足够了。开始坐标总是小于结束坐标。
一支弓箭可以沿着 x 轴从不同点完全垂直地射出。在坐标 x 处射出一支箭，若有一个气球的直径的开始和结束坐标为 xstart，xend， 且满足 xstart ≤ x ≤ xend，则该气球会被引爆。可以射出的弓箭的数量没有限制。 弓箭一旦被射出之后，可以无限地前进。我们想找到使得所有气球全部被引爆，所需的弓箭的最小数量。
给你一个数组 points ，其中 points [i] = [xstart,xend] ，返回引爆所有气球所必须射出的最小弓箭数。
示例 :
示例 1：
输入：points = [[10,16],[2,8],[1,6],[7,12]]
输出：2
解释：对于该样例，x = 6 可以射爆 [2,8],[1,6] 两个气球，以及 x = 11 射爆另外两个气球
示例 2：
输入：points = [[1,2],[3,4],[5,6],[7,8]]
输出：4
示例 3：
输入：points = [[1,2],[2,3],[3,4],[4,5]]
输出：2
示例 4：
输入：points = [[1,2]]
输出：1
示例 5：
输入：points = [[2,3],[2,3]]
输出：1
提示：
0 <= points.length <= 10^4
points[i].length == 2
-2^31 <= xstart < xend <= 2^31 - 1
思路:
贪心算法
对区间的右端点排序
至少需要一支箭
每次从最右射出一支箭, 如果区间超出当前区间, 更新区间, 加一支箭
时间复杂度O(nlgn), 空间复杂度O(1)
代码:
C++:
class Solution
{
public:
int findMinArrowShots(vector>& points)
{
if (points.empty()) return 0;
sort(points.begin(), points.end(), [](const vector& a, const vector& b) { return a[0] < b[0]; });
int result = 1;
for (int i = 1; i < points.size(); i++)
{
if (points[i][0] > points[i - 1][1]) ++result;
else points[i][1] = min(points[i - 1][1], points[i][1]);
}
return result;
}
};
Java:
class Solution {
public int findMinArrowShots(int[][] points) {
if (points.length == 0) return 0;
Arrays.sort(points, (a, b) -> a[1] < b[1] ? -1 : 1);
int result = 1, cur = points[0][1];
for (int i = 1; i < points.length; i++) if (cur < points[i][0]) {
++result;
cur = points[i][1];
}
return result;
}
}
Python:
class Solution:
def findMinArrowShots(self, points: List[List[int]]) -> int:
if not points:
return 0
points.sort(key=lambda p: p[1])
result, cur = 1, points[0][1]
for point in points[1:]:
if cur < point[0]:
result += 1
cur = point[1]
return result