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2021.Brightest-Position-on-Street

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2021. Brightest Position on Street

题目

A perfectly straight street is represented by a number line. The street has street lamp(s) on it and is represented by a 2D integer array lights. Each lights[i] = [positioni, rangei] indicates that there is a street lamp at position positioni that lights up the area from [positioni - rangei, positioni + rangei] (inclusive).

The brightness of a position p is defined as the number of street lamp that light up the position p.

Given lights, return the brightest position on the street. If there are multiple brightest positions, return the smallest one.

Example 1:

https://assets.leetcode.com/uploads/2021/09/28/image-20210928155140-1.png

Input: lights = [[-3,2],[1,2],[3,3]] Output: -1 Explanation: The first street lamp lights up the area from [(-3) - 2, (-3) + 2] = [-5, -1]. The second street lamp lights up the area from [1 - 2, 1 + 2] = [-1, 3]. The third street lamp lights up the area from [3 - 3, 3 + 3] = [0, 6]. Position -1 has a brightness of 2, illuminated by the first and second street light. Positions 0, 1, 2, and 3 have a brightness of 2, illuminated by the second and third street light. Out of all these positions, -1 is the smallest, so return it.

Example 2:

Input: lights = [[1,0],[0,1]] Output: 1 Explanation: The first street lamp lights up the area from [1 - 0, 1 + 0] = [1, 1]. The second street lamp lights up the area from [0 - 1, 0 + 1] = [-1, 1]. Position 1 has a brightness of 2, illuminated by the first and second street light. Return 1 because it is the brightest position on the street.

Example 3:

Input: lights = [[1,2]] Output: -1 Explanation: The first street lamp lights up the area from [1 - 2, 1 + 2] = [-1, 3]. Positions -1, 0, 1, 2, and 3 have a brightness of 1, illuminated by the first street light. Out of all these positions, -1 is the smallest, so return it.

Constraints:

  • 1 <= lights.length <= 105
  • lights[i].length == 2
  • 108 <= positioni <= 108
  • 0 <= rangei <= 108

题目大意

一条完全笔直的街道由一条数字线表示。街道上有路灯，由二维数据表示。每个 lights[i] = [positioni, rangei] 表示位置 i 处有一盏路灯，灯可以照亮从 [positioni - rangei, positioni + rangei] （含）的区域。 位置 p 的亮度定义为点亮位置 p 的路灯数量。 给定路灯，返回街道上最亮的位置。如果有多个最亮的位置，则返回最小的一个。

解题思路

  • 先将每个路灯的起始和终点位置计算出来。这样我们得到了一堆坐标点。假设灯照亮的范围是 [A, B]，那么在坐标轴上 A 坐标点处 + 1， B + 1 坐标点处 -1 。这样处理的含义是：坐标点 A 可以被一盏灯照亮，所以它照亮次数加一，坐标点 B + 1 出了灯照亮的范围了，所以照亮次数减一。那么从坐标轴坐标开始扫一遍，每次遇到 + 1 的时候就 + 1，遇到 - 1 的地方就 - 1。如此可以算出某个坐标点处，可以被灯照亮的总次数。
  • 需要注意的点是，题目给的测试数据可能会有单点照亮的情况，即某一盏灯只照亮一个坐标点，灯照范围为 0。同一个坐标点也可能是多个灯的起点。用一个 map 去重坐标点即可。

代码

package leetcode import ( "sort" ) typelightItemstruct { indexintsignint } funcbrightestPosition(lights [][]int) int { lightMap, lightItems:=map[int]int{}, []lightItem{} for_, light:=rangelights { lightMap[light[0]-light[1]] +=1lightMap[light[0]+light[1]+1] -=1 } fork, v:=rangelightMap { lightItems=append(lightItems, lightItem{index: k, sign: v}) } sort.SliceStable(lightItems, func(i, jint) bool { returnlightItems[i].index<lightItems[j].index }) res, border, tmp:=0, 0, 0for_, v:=rangelightItems { tmp+=v.signifborder<tmp { res=v.indexborder=tmp } } returnres }
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