[Swift]LeetCode1033. 边框着色 | Moving Stones Until Consecutive

Given a 2-dimensional grid of integers, each value in the grid represents the color of the grid square at that location.

Two squares belong to the same connected component if and only if they have the same color and are next to each other in any of the 4 directions.

The border of a connected component is all the squares in the connected component that are either 4-directionally adjacent to a square not in the component, or on the boundary of the grid (the first or last row or column).

Given a square at location (r0, c0) in the grid and a color, color the border of the connected component of that square with the given color, and return the final grid.

Example 1:

Input: grid = [[1,1],[1,2]], r0 = 0, c0 = 0, color = 3
Output: [[3, 3], [3, 2]]

Example 2:

Input: grid = [[1,2,2],[2,3,2]], r0 = 0, c0 = 1, color = 3
Output: [[1, 3, 3], [2, 3, 3]]

Example 3:

Input: grid = [[1,1,1],[1,1,1],[1,1,1]], r0 = 1, c0 = 1, color = 2
Output: [[2, 2, 2], [2, 1, 2], [2, 2, 2]]

Note:

1. 1 <= grid.length <= 50
2. 1 <= grid[0].length <= 50
3. 1 <= grid[i][j] <= 1000
4. 0 <= r0 < grid.length
5. 0 <= c0 < grid[0].length
6. 1 <= color <= 1000

只有当两个网格块的颜色相同，而且在四个方向中任意一个方向上相邻时，它们属于同一连通分量。

连通分量的边界是指连通分量中的所有与不在分量中的正方形相邻（四个方向上）的所有正方形，或者在网格的边界上（第一行/列或最后一行/列）的所有正方形。

给出位于 (r0, c0) 的网格块和颜色 color，使用指定颜色 color 为所给网格块的连通分量的边界进行着色，并返回最终的网格 grid 。

示例 1：

输入：grid = [[1,1],[1,2]], r0 = 0, c0 = 0, color = 3
输出：[[3, 3], [3, 2]]

示例 2：

输入：grid = [[1,2,2],[2,3,2]], r0 = 0, c0 = 1, color = 3
输出：[[1, 3, 3], [2, 3, 3]]

示例 3：

输入：grid = [[1,1,1],[1,1,1],[1,1,1]], r0 = 1, c0 = 1, color = 2
输出：[[2, 2, 2], [2, 1, 2], [2, 2, 2]]

提示：

1. 1 <= grid.length <= 50
2. 1 <= grid[0].length <= 50
3. 1 <= grid[i][j] <= 1000
4. 0 <= r0 < grid.length
5. 0 <= c0 < grid[0].length
6. 1 <= color <= 1000

Memory Usage: 19.1 MB

 1 class Solution {
 2     func numMovesStones(_ a: Int, _ b: Int, _ c: Int) -> [Int] {
 3         var arr:[Int] = [a,b,c].sorted(by:<)
 4         let a = arr[0]
 5         let b = arr[1]
 6         let c = arr[2]
 7         var minNum:Int = 0
 8         if abs(a - b) == 2 || abs(c - b) == 2
 9         {
10             minNum +=  1
11         }
12         else
13         {
14             if abs(a - b) > 2
15             {
16                 minNum +=  1
17             }
18             if  abs(c - b) > 2
19             {
20                 minNum +=  1
21             }
22         }
23         return [minNum, abs(b - a) + abs(c - b) - 2]
24     }
25 }

原文地址：https://www.cnblogs.com/strengthen/p/10783233.html