[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 = [1,1,2,1,3,2,2,3,3], queries = [[1,3],[2,2],[6,1]]
Output: [3,0,3]
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 = [1,2], 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 上进行一些查询操作 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 = [1,1,2,1,3,2,2,3,3], queries = [[1,3],[2,2],[6,1]]
输出：[3,0,3]
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 = [1,2], queries = [[0,3]]
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

class Solution {
    func shortestDistanceColor(_ colors: [Int], _ queries: [[Int]]) -> [Int] {
        let n:Int = colors.count
        var left:[[Int]] = [[Int]](repeating:[Int](repeating:-1,count:n),count:4)
        var right:[[Int]] = [[Int]](repeating:[Int](repeating:-1,count:n),count:4)
        for shade in 1...3
        {
            if colors[0] == shade
            {
                left[shade][0] = 0
            }
            for i in 1..<n
            {
                if left[shade][i-1] != -1
                {
                    left[shade][i] = left[shade][i-1] + 1
                }
                if colors[i] == shade
                {
                    left[shade][i] = 0
                }
            }
        }
        for shade in 1...3
        {
            if colors[n-1] == shade
            {
                right[shade][n-1] = 0
            }
            for i in stride(from:n - 2,through:0,by:-1)
            {
                if right[shade][i+1] != -1
                {
                    right[shade][i] = right[shade][i+1] + 1
                }
                if colors[i] == shade
                {
                    right[shade][i] = 0
                }
            }
        }
        var result:[Int] = [Int]()
        for query in queries
        {
            var index:Int = query[0]
            var req_color = query[1]
            var x:Int = left[req_color][index]
            var y:Int = right[req_color][index]
            var ans:Int = 0
            if x == -1 || y == -1
            {
                ans = max(x,y)
            }
            else
            {
                ans = min(x,y)
            }
            result.append(ans)
        }
        return result
    }
}