[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

	class Solution {
	fun minJumps(arr: IntArray): Int {
	val map = mutableMapOf<Int,MutableList<Int>>()
	for(i in 1 until arr.size) {
	if(map[arr[i]] == null) {
	map[arr[i]] = mutableListOf<Int>(i)
	} else {
	map[arr[i]]!!.add(i)
	}
	}
	var queue = mutableListOf<List<Int>>()
	queue.add(listOf(0))
	var visited = mutableSetOf<Int>()
	visited.add(0)
	var ans = 0
	while(queue.isNotEmpty()) {
	val paths = queue.first()
	queue.removeAt(0)
	val nextPath = mutableListOf<Int>()
	for(index in paths) {
	if(index == arr.lastIndex) {
	return ans
	}
	if(map[arr[index]] != null) {
	for(i in map[arr[index]]!!) {
	if(!visited.contains(i)) {
	visited.add(i)
	nextPath.add(i)
	}
	}
	map.remove(arr[index])
	}
	if(index > 0 && !visited.contains(index-1)) {
	visited.add(index-1)
	nextPath.add(index-1)
	}
	if(index+1 < arr.size && !visited.contains(index+1)) {
	visited.add(index+1)
	nextPath.add(index+1)
	}
	}
	queue.add(nextPath)
	++ans
	}
	return ans
	}
	}

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