Leetcode simple question: find the nearest point with the same X or Y coordinate

subject

Here are two integers x and y , It means that you are in a Cartesian coordinate system (x, y) It's about . meanwhile , Give you an array in the same coordinate system points , among points[i] = [ai, bi] It means that (ai, bi) There's a point in the middle . When a point has the same... As where you are x Coordinates or the same y Coordinates , We call this point Effective .
Please return to your current position Manhattan distance Current It works Subscript of point （ Subscript from 0 Start ）. If there are multiple nearest valid points , Please return to subscript Minimum One of the . If it doesn't work , Please return -1 .
Two points (x1, y1) and (x2, y2) Between Manhattan distance by abs(x1 - x2) + abs(y1 - y2)
Example 1：
Input ：x = 3, y = 4, points = [[1,2],[3,1],[2,4],[2,3],[4,4]]
Output ：2
explain ： Of all the points ,[3,1],[2,4] and [4,4] It's effective . In effect ,[2,4] and [4,4] Manhattan is the closest to your current location , All for 1 .[2,4] The lowest subscript of , So back 2 .
Example 2：
Input ：x = 3, y = 4, points = [[3,4]]
Output ：0
Tips ： The answer can be the same as your current position .
Example 3：
Input ：x = 3, y = 4, points = [[2,3]]
Output ：-1
explain ： No, Effective point .
Tips ：
1 <= points.length <= 10^4
points[i].length == 2
1 <= x, y, ai, bi <= 10^4
source ： Power button （LeetCode）

Their thinking

   According to the meaning of the topic , Find the effective point first . The definition of effective point is points The element of has one or two with a given x and y It corresponds to equality , Then the current element is a valid point . Then calculate the distance between the given point and the effective point , Count the nearest valid point and return the subscript .

class Solution:
    def nearestValidPoint(self, x: int, y: int, points: List[List[int]]) -> int:
        distance=float('inf')
        index=-1
        for i in range(len(points)):
            if x==points[i][0] or y==points[i][1]:
                if distance>abs(points[i][0]-x)+abs(points[i][1]-y):
                    distance=abs(points[i][0]-x)+abs(points[i][1]-y)
                    index=i
        return index