[Swift]LeetCode1182. 与目标颜色间的最短距离 ｜ Shortest Distance to Target Color

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You are given an array colors,in which there are three colors: 1, 2 and 3.

You are also given some querIEs. Each query consists of two integers i and c,return the shortest distance between the given index i and the target color c. If there is no solution return -1.

 

Example 1:

input: colors = [1,1,2,3,3],querIEs = [[1,[2,2],[6,1]]Output: [3,3]Explanation: The nearest 3 from index 1 is at index 4 (3 steps away).The nearest 2 from index 2 is at index 2 itself (0 steps away).The nearest 1 from index 6 is at index 3 (3 steps away).

Example 2:

input: colors = [1,querIEs = [[0,3]]Output: [-1]Explanation: There is no 3 in the array.

 

Constraints:

1 <= colors.length <= 5*10^4 1 <= colors[i] <= 3 1 <= querIEs.length <= 5*10^4 querIEs[i].length == 2 0 <= querIEs[i][0] < colors.length 1 <= querIEs[i][1] <= 3

 

给你一个数组 colors，里面有  1、2、 3 三种颜色。

我们需要在 colors 上进行一些查询 *** 作 querIEs，其中每个待查项都由两个整数 i 和 c 组成。

现在请你帮忙设计一个算法，查找从索引 i 到具有目标颜色 c 的元素之间的最短距离。

如果不存在解决方案，请返回 -1。

 

示例 1：

输入：colors = [1,1]]输出：[3,3]解释： 距离索引 1 最近的颜色 3 位于索引 4（距离为 3）。距离索引 2 最近的颜色 2 就是它自己（距离为 0）。距离索引 6 最近的颜色 1 位于索引 3（距离为 3）。

示例 2：

输入：colors = [1,3]]输出：[-1]解释：colors 中没有颜色 3。

 

提示：

1 <= colors.length <= 5*10^4 1 <= colors[i] <= 3 1 <= querIEs.length <= 5*10^4 querIEs[i].length == 2 0 <= querIEs[i][0] < colors.length 1 <= querIEs[i][1] <= 3

Runtime: 1880 ms

Memory Usage: 28.9 MB

 1 class Solution { 2     func shortestdistancecolor(_ colors: [Int],_ querIEs: [[Int]]) -> [Int] { 3         let n:Int = colors.count 4         var left:[[Int]] = [[Int]](repeating:[Int](repeating:-1,count:n),count:4) 5         var right:[[Int]] = [[Int]](repeating:[Int](repeating:-1,count:4) 6         for shade in 1...3 7         { 8             if colors[0] == shade 9             {10                 left[shade][0] = 011             }12             for i in 1..<n13             {14                 if left[shade][i-1] != -115                 {16                     left[shade][i] = left[shade][i-1] + 117                 }18                 if colors[i] == shade19                 {20                     left[shade][i] = 021                 }22             }23         }24         for shade in 1...325         {26             if colors[n-1] == shade27             {28                 right[shade][n-1] = 029             }30             for i in strIDe(from:n - 2,through:0,by:-1)31             {32                 if right[shade][i+1] != -133                 {34                     right[shade][i] = right[shade][i+1] + 135                 }36                 if colors[i] == shade37                 {38                     right[shade][i] = 039                 }40             }41         }42         var result:[Int] = [Int]()43         for query in querIEs44         {45             var index:Int = query[0]46             var req_color = query[1]47             var x:Int = left[req_color][index]48             var y:Int = right[req_color][index]49             var ans:Int = 050             if x == -1 || y == -151             {52                 ans = max(x,y)53             }54             else55             {56                 ans = min(x,y)57             }58             result.append(ans)59         }60         return result61     }62 }

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