greedy——452. Minimum Number of Arrows to Burst Balloons［medium］

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. 

Example:

Input:[[10,16], [2,8], [1,6], [7,12]]Output:2Explanation: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).

抽象出来的意思是，选取一个数，让它经过最多的区间数，问这样的数至少有几个。

解题思路

贪心算法，局部最优解 ＝ 整体最优解，每次选一个经过最大区间的数（其实不用确定这个数，只要计算重叠区间数，所有重叠区间都可以用一个数穿过），一直到经过所有区间为止。解题示意图如下：

从图中得到算法如下：

1、给pair排序，默认排序就是按第一个大小，第一个相同再比较第二个，升序排好

2、遍历数组，用tmp（第i个数组元素）表示当前区间，检查当前区间与检查到的区间是否有重叠，重叠的话，tmp ＝ 重叠部分，一直进行下去

3、如果无重叠，则tmp ＝ 检查到的区间，表示arrow的计数器＋＋，重复2

代码如下

class Solution {public:  bool inter(pair<int, int> a, pair<int, int> b) {if (a.second < b.first || b.second < a.first)return false;return true;}int findMinArrowShots(vector<pair<int, int>>& points) {if (points.empty())return 0;sort(points.begin(), points.end());int count = 0;for (int i = 0; i < points.size(); i++) {pair<int, int> tmp = points[i];int j = i + 1;for (j; j < points.size(); j++) {if (inter(tmp, points[j])) {tmp.first = fmax(tmp.first, points[j].first);tmp.second = fmin(tmp.second, points[j].second);continue;}break;}i = j - 1;count++;}return count;}};

inter函数判断是否有交集。

fmax返回两者较大者，fmin相反。