leecode刷题:1779. Find Nearest Point That Has the Same X or Y Coordinate

in hive-180932 •  7 months ago 

You are given two integers, x and y, which represent your current location on a Cartesian grid: (x, y). You are also given an array points where each points[i] = [ai, bi] represents that a point exists at (ai, bi). A point is valid if it shares the same x-coordinate or the same y-coordinate as your location.

Return the index (0-indexed) of the valid point with the smallest Manhattan distance from your current location. If there are multiple, return the valid point with the smallest index. If there are no valid points, return -1.

The Manhattan distance between two points (x1, y1) and (x2, y2) is 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
Explanation: Of all the points, only [3,1], [2,4] and [4,4] are valid. Of the valid points, [2,4] and [4,4] have the smallest Manhattan distance from your current location, with a distance of 1. [2,4] has the smallest index, so return 2.
Example 2:

Input: x = 3, y = 4, points = [[3,4]]
Output: 0
Explanation: The answer is allowed to be on the same location as your current location.
Example 3:

Input: x = 3, y = 4, points = [[2,3]]
Output: -1
Explanation: There are no valid points.

Constraints:

1 <= points.length <= 104
points[i].length == 2
1 <= x, y, ai, bi <= 104

这是周赛的题,说实话英文题目看不懂。我用例子刷子好多回才看懂了。

用的方法很简单粗暴,就是先判断是否有x或者y相等的,然后计算出距离long,之后判断long是否最小,最后输出long。

class Solution:
    def nearestValidPoint(self, x: int, y: int, points: List[List[int]]) -> int:
        mins=8888
        for i in range(len(points)):
            long = 9999
            if points[i][0] == x or points[i][1] == y:
                long =  abs(x - points[i][0]) + abs(y - points[i][1])
            if long < mins:
                mins=long
                re=i
        try:
            return re
        except:
            return -1
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