LeetCode452. Minimum Number of Arrows to Burst Balloons

文章目录

一、题目

There are some spherical balloons taped onto a flat wall that represents the XY-plane. The balloons are represented as a 2D integer array points where points[i] = [xstart, xend] denotes a balloon whose horizontal diameter stretches between xstart and xend. You do not know the exact y-coordinates of the balloons.

Arrows can be shot up directly vertically (in the positive y-direction) from different points along the x-axis. A balloon with xstart and xend is burst by an arrow shot at x if xstart <= x <= xend. There is no limit to the number of arrows that can be shot. A shot arrow keeps traveling up infinitely, bursting any balloons in its path.

Given the array points, 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: The balloons can be burst by 2 arrows:

  • Shoot an arrow at x = 6, bursting the balloons [2,8] and [1,6].
  • Shoot an arrow at x = 11, bursting the balloons [10,16] and [7,12].
    Example 2:

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

Output: 4

Explanation: One arrow needs to be shot for each balloon for a total of 4 arrows.

Example 3:

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

Output: 2

Explanation: The balloons can be burst by 2 arrows:

  • Shoot an arrow at x = 2, bursting the balloons [1,2] and [2,3].
  • Shoot an arrow at x = 4, bursting the balloons [3,4] and [4,5].

Constraints:

1 <= points.length <= 105

points[i].length == 2

-231 <= xstart < xend <= 231 - 1

二、题解

cpp 复制代码
class Solution {
public:
    static bool cmp(vector<int>& a,vector<int>& b){
        return a[0] < b[0];
    }
    int findMinArrowShots(vector<vector<int>>& points) {
        int n = points.size();
        //按左边界从小到大排序
        sort(points.begin(),points.end(),cmp);
        int res = 1;
        for(int i = 1;i < n;i++){
            //如果当前气球左边界大于上一个气球的右边界
            if(points[i][0] > points[i-1][1]) res++;
            //如果重合,则更新右端点值
            else points[i][1] = min(points[i][1],points[i-1][1]);
        }
        return res;
    }
};
相关推荐
Tisfy1 小时前
LeetCode 3240.最少翻转次数使二进制矩阵回文 II:分类讨论
算法·leetcode·矩阵·题解·回文·分类讨论
橘子遇见BUG2 小时前
算法日记 31 day 动态规划(01背包)
算法·动态规划
东方巴黎~Sunsiny2 小时前
java-图算法
java·开发语言·算法
ac-er88882 小时前
PHP二维数组排序算法函数
算法·php·排序算法
Tisfy3 小时前
LeetCode 3244.新增道路查询后的最短距离 II:贪心(跃迁合并)-9行py(O(n))
算法·leetcode·题解·贪心·思维
DdddJMs__1353 小时前
C语言 | Leetcode C语言题解之第564题寻找最近的回文数
c语言·leetcode·题解
Matlab程序猿小助手4 小时前
【MATLAB源码-第218期】基于matlab的北方苍鹰优化算法(NGO)无人机三维路径规划,输出做短路径图和适应度曲线.
开发语言·嵌入式硬件·算法·matlab·机器人·无人机
qq_428639614 小时前
植物明星大乱斗15
c++·算法·游戏
捕鲸叉4 小时前
C++创建型模式之生成器模式
开发语言·c++·建造者模式
sxtyjty4 小时前
人机打怪小游戏(非常人机)
c++