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# 扫描线 | ||
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--- | ||
数飞机举例: | ||
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### 扫描线定义: | ||
不需要检测每一时刻, 只需要检测起点或者终点的位置(变化的位置只有起点或者终点) | ||
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[[数飞机]]举例: | ||
``` | ||
暴力扫描: | ||
遍历每个时刻, 检测每个时刻有多少飞机 | ||
扫描线: | ||
不需要检测每一时刻, 只需要检测起点或者终点的位置(交点变化的位置只有起点或者终点) | ||
不需要检测每一个时刻,因为只有当飞机起飞或者降落的时候,它会对这个天上飞机的数量有影响。 | ||
我们就只要扫描这些起飞或者降落的时间就可以大幅度降低复杂度 | ||
``` | ||
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### 题目 | ||
[[数飞机]] | ||
[[最大线段重合问题]] | ||
[[拼车]] | ||
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[H] | ||
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#单调栈 | ||
#数组压缩 | ||
#德高望重的噩梦题 | ||
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--- | ||
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# 拼车 | ||
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#扫描线 | ||
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1094.拼车 | ||
[M] | ||
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--- | ||
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车上最初有 capacity 个空座位。车 只能 向一个方向行驶(也就是说,不允许掉头或改变方向) | ||
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给定整数 capacity 和一个数组 trips , `trip[i] = [numPassengersi, fromi, toi]` 表示第 i 次旅行有 numPassengersi 乘客,接他们和放他们的位置分别是 fromi 和 toi 。这些位置是从汽车的初始位置向东的公里数。 | ||
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当且仅当你可以在所有给定的行程中接送所有乘客时,返回 true,否则请返回 false。 | ||
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``` | ||
示例 1: | ||
输入:trips = [[2,1,5],[3,3,7]], capacity = 4 | ||
输出:false | ||
示例 2: | ||
输入:trips = [[2,1,5],[3,3,7]], capacity = 5 | ||
输出:true | ||
提示: | ||
1 <= trips.length <= 1000 | ||
trips[i].length == 3 | ||
1 <= numPassengersi <= 100 | ||
0 <= fromi < toi <= 1000 | ||
1 <= capacity <= 10^5 | ||
``` |
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# 数飞机 | ||
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391.数飞机 | ||
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[M] | ||
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#扫描线 | ||
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--- | ||
https://www.lintcode.com/problem/391/ | ||
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描述 | ||
给出飞机的起飞和降落时间的列表,用序列 interval 表示. 请计算出天上同时最多有多少架飞机? | ||
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如果多架飞机降落和起飞在同一时刻,我们认为降落有优先权。 | ||
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``` | ||
样例 | ||
样例 1: | ||
输入: [(1, 10), (2, 3), (5, 8), (4, 7)] | ||
输出: 3 | ||
解释: | ||
第一架飞机在1时刻起飞, 10时刻降落. | ||
第二架飞机在2时刻起飞, 3时刻降落. | ||
第三架飞机在5时刻起飞, 8时刻降落. | ||
第四架飞机在4时刻起飞, 7时刻降落. | ||
在5时刻到6时刻之间, 天空中有三架飞机. | ||
样例 2: | ||
输入: [(1, 2), (2, 3), (3, 4)] | ||
输出: 1 | ||
解释: 降落优先于起飞. | ||
``` | ||
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ref: [[最大线段重合问题]] |
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# 最大线段重合问题-堆实现 | ||
# 最大线段重合问题 | ||
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#堆 | ||
#优先队列 | ||
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--- | ||
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4 changes: 2 additions & 2 deletions
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.../200. ★算法题目汇总★/★算法面试题汇总/最大线段重叠问题-线段树实现.md → .../200. ★算法题目汇总★/★算法面试题汇总/最大线段重合问题-线段树实现.md
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# 最大线段重叠问题 | ||
# 最大线段重合问题 | ||
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#线段树 | ||
#优先队列 | ||
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#堆 | ||
#扫描线 | ||
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--- | ||
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