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
}