1345. Jump Game IV

Hard

Given an array of integers arr, you are initially positioned at the first index of the array.

In one step you can jump from index i to index:

• i + 1 where: i + 1 < arr.length.

• i - 1 where: i - 1 >= 0.

• j where: arr[i] == arr[j] and i != j.

Return the minimum number of steps to reach the last index of the array.

Notice that you can not jump outside of the array at any time.

Example 1:

Input: arr = [100,-23,-23,404,100,23,23,23,3,404]
Output: 3
Explanation: You need three jumps from index 0 --> 4 --> 3 --> 9. Note that index 9 is the last index of the array.

Example 2:

Input: arr = [7]
Output: 0
Explanation: Start index is the last index. You do not need to jump.

Example 3:

Input: arr = [7,6,9,6,9,6,9,7]
Output: 1
Explanation: You can jump directly from index 0 to index 7 which is last index of the array.

Constraints:

• 1 <= arr.length <= 5 * 10^4

• -10^8 <= arr[i] <= 10^8

解題

BFS

func minJumps(arr []int) int {
    steps := 0
    n := len(arr)   

    m := make(map[int][]int) 
    for i:=0; i<n; i++ {
        if _, ok := m[arr[i]]; !ok {
            m[arr[i]] = []int{ i }
        } else {
            m[arr[i]] = append(m[arr[i]], i)
        }
    }

    visited := make([]bool, n)
    visited[0] = true

    visitedv := make(map[int]bool) // 經過的數字，已經進過queue了

    queue := make([]int, 0) 
    queue = append(queue, 0)

    for len(queue) != 0 {
        newqueue := make([]int, 0)
        size := len(queue)
        for i:=0; i<size; i++ {
            current := queue[i]

            if current == n-1 { return steps }

            canJump := []int{} // 當前數字可以跳去的位置儲存

            if !visitedv[arr[current]] { // 有些位置已經去過，就不加入下回的queue了，不做這一步會 TLE
                canJump = m[arr[current]]
            }

            canJump = append(canJump, current + 1)
            canJump = append(canJump, current - 1)

            for _, v := range canJump {
                if v > 0 && v < n && !visited[v] {
                    visited[v] = true
                    newqueue = append(newqueue, v)
                }
            }

            visitedv[arr[current]] = true
        }
        queue = newqueue //進入下一回合，step也要加一
        steps++
    }

    return -1 
}