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描述
You are given a 2D integer array stockPrices where stockPrices[i] = [dayi, pricei] indicates the price of the stock on day dayi is pricei. A line chart is created from the array by plotting the points on an XY plane with the X-axis representing the day and the Y-axis representing the price and connecting adjacent points. One such example is shown below:
Example 1:
Input: stockPrices = [[1,7],[2,6],[3,5],[4,4],[5,4],[6,3],[7,2],[8,1]]
Output: 3
Explanation:
The diagram above represents the input, with the X-axis representing the day and Y-axis representing the price.
The following 3 lines can be drawn to represent the line chart:
- Line 1 (in red) from (1,7) to (4,4) passing through (1,7), (2,6), (3,5), and (4,4).
- Line 2 (in blue) from (4,4) to (5,4).
- Line 3 (in green) from (5,4) to (8,1) passing through (5,4), (6,3), (7,2), and (8,1).
It can be shown that it is not possible to represent the line chart using less than 3 lines.
Example 2:
Input: stockPrices = [[3,4],[1,2],[7,8],[2,3]]
Output: 1
Explanation:
As shown in the diagram above, the line chart can be represented with a single line.
Note:
1 <= stockPrices.length <= 10^5
stockPrices[i].length == 2
1 <= dayi, pricei <= 10^9
All dayi are distinct.
解析
根据题意，给定一个二维整数数组 stockPrices ，其中 stockPrices[i] = [dayi, pricei] 表示第 i 天的股票价格为 pricei 。 通过在 XY 平面上绘制点来按照数组数据创建折线图，其中 X 轴代表当天，Y 轴代表价格并连接相邻点。 返回将所有点都连接起来所使用的直线最小数量。
我们分析题目就能发现，能减少直线数量的关键就在于，相邻的三个点中前两个点和后两个点能不能是斜率一样的，如果能那么就不用增加直接的数量，否则斜率不一样的情况下肯定要增加不用的直线数量。
比赛的时候我有点疯了，直接按照斜率公式去判断两个斜率是否相等，并给出了斜率相等的容忍误差，但是这仍然无法满足题意，因为题目中的数都太大了，一般的容忍误差无法做到。其实最简单的方法就是将原始的公式
dy / dx == dyy / dxx
变化为，这样就能避免精度损失：
dy * dxx == dyy * dx
时间复杂度为 O(N) ，空间复杂度为 O(1) 。
解答
class Solution(object):
def minimumLines(self, stockPrices):
"""
:type stockPrices: List[List[int]]
:rtype: int
"""
N = len(stockPrices)
if N == 1: return 0
if N == 2: return 1
result = 1
stockPrices.sort()
for i in range(1, N):
if -1<i-1 and i+1<N:
dx = stockPrices[i][0] - stockPrices[i-1][0]
dy = stockPrices[i][1] - stockPrices[i-1][1]
dxx = stockPrices[i+1][0] - stockPrices[i][0]
dyy = stockPrices[i+1][1] - stockPrices[i][1]
if dy*dxx == dyy * dx:
continue
else:
result += 1
return result
运行结果
79 / 79 test cases passed.
Status: Accepted
Runtime: 2340 ms
Memory Usage: 61.6 MB
原题链接
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