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.

?

easy 題目，先確認(rèn)數(shù)據(jù)是否是有效的，然后再計(jì)算到位置點(diǎn)的曼哈頓距離，保留一個(gè)最小值，然后返回即可。

Constraints:

• 1 <= points.length <= 104

• points[i].length == 2

• 1 <= x, y, ai, bi?<= 104

Runtime:?2 ms, faster than?64.54%?of?Java?online submissions for?Find Nearest Point That Has the Same X or Y Coordinate.

Memory Usage:?50.7 MB, less than?12.67%?of?Java?online submissions for?Find Nearest Point That Has the Same X or Y Coordinate.