[LeetCode] 1345. Jump Game IV

轉自LeetCode

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 don't 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.

Example 4:

Input: arr = [6,1,9]
Output: 2

Example 5:

Input: arr = [11,22,7,7,7,7,7,7,7,22,13]
Output: 3

 

Constraints:

• 1 <= arr.length <= 5 * 104
• -108 <= arr[i] <= 108

Solution

1306. Jump Game III 的衍生題

除了正常的可以前後移動一個格的條件外

多了一個新條件，i can move to j when arr[i] = arr[j] && i != j

另外，題目要求的回傳不是是不是可以到達最後一個位置，而是最少的步數

一樣用 BFS 來解，但 queue 存的變成是 list of index

因為同一個步數下，可以有多個選擇

至於新條件的跳法，用一個 map 的方式來儲存，用過之後就刪掉

不刪掉會 TLE，因為會重複計算

kotlin