LeetCode 452. Minimum Number of Arrows to Burst Balloons 题解（C++）

LeetCode 452. Minimum Number of Arrows to Burst Balloons 题解（C++）

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题目描述

• There are a number of 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. Start is always smaller than end. There will be at most 104 balloons.
• 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 travelling up infinitely. The problem is to find the minimum number of arrows that must be shot to burst all balloons.

举例

• Input:
  [[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).

思路

• 首先自己定义一个比较函数，将输入的pair数组先按照pair的first升序排列，若pair的first相等，则按照pair的second升序排列；
• 首先将排列后的数组的第一个元素的second保存在right，遍历排序后的数组，若数组元素的first小于等于right，则表示该元素的某些范围在right之内，可以被一枪射中，此时将right更新为该元素的second和right中较小的一个；若数组元素的first大于right，则该气球不能与前面的气球一起射中，此时更新right为当前元素的second，并使射击次数result加1。

代码

我的代码

class Solution 
{
public:
    static bool mySort(pair<int, int> a, pair<int, int> b)
    {
        if (a.first == b.first)
        {
            return a.second < b.second;
        }
        return a.first < b.first;
    }
    int findMinArrowShots(vectorint, int>>& points) 
    {
        if (points.empty())
        {
            return 0;
        }   
        sort(points.begin(), points.end(), mySort);
        int result = 1;
        int right = points[0].second;
        for (int i = 1; i < points.size(); ++i)
        {
            pair<int, int> temp = points[i];
            if (temp.first > right)
            {
                ++result;
                right = temp.second;
            }
            else
            {
                right = min(right, temp.second);
            }
        }
        return result;
    }
};