[Leetcode] 452. Minimum Number of Arrows to Burst Balloons 解题报告

题目：

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.

Example:

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：

采用贪心的策略，仅仅在不得不打爆某个气球的时候，我们才发射一个arrow。具体思路是：首先将Balloon按照x_start进行排序，然后依次遍历，并不断更新x_end的最小值。一旦发现当前遍历的气球的x_start已经大于目前x_end的最小值了，说明我们如果不在x_end最小值处发射一个arrow，则会导致某些气球在后面不再有机会被打爆，所以此时我们就发射一个arrow（此时arrow发射的位置在x_end的最小值处）。值得注意的是：最后还需要发射一个arrow，打掉剩余的气球。

虽然遍历的时间复杂度是O(n)，但是由于涉及排序，所以算法的总体时间复杂度是O(nlogn)，空间复杂度是O(1)。

代码：

class Solution {
public:
    int findMinArrowShots(vector<pair<int, int>>& points) {
        if (points.size() == 0) {
            return 0;
        }
        sort(points.begin(), points.end(), BalloonComp);
        int ret = 0, end = INT_MAX;
        for (auto point : points) {
            if (point.first > end) {
                ++ret;
                end = point.second;
            }
            else {
                end = min(end, point.second);
            }
        }
        return ++ret;
    }
private:
    struct BalloonCompare {
        bool operator() (const pair<int, int> &a, const pair<int, int> &b) {
            return a.first < b.first;
        }
    } BalloonComp;
};