1345. Jump Game IV (H)

题目

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 * 10^4
• -10^8 <= arr[i] <= 10^8

题意

给定一个数组，对应其中一个下标i，下一步可以到达的下标为i-1、i+1、所有满足arr[i]==arr[j]的j。要求能从0到达最后一个下标的最小步数。

思路

按照提示，将数组重构成一张表，结点的值是数组中的下标，且与前后两个下标对应的结点、所有与该结点在数组中对应的值相等的结点相连。用BFS找最短距离。

代码实现

Java

class Solution {
    public int minJumps(int[] arr) {
        if (arr.length == 1) {
            return 0;
        }
        Map<Integer, List<Integer>> inverse = new HashMap<>();
        for (int i = 0; i < arr.length; i++) {
            inverse.putIfAbsent(arr[i], new ArrayList<>());
            inverse.get(arr[i]).add(i);
        }
        int steps = 0;
        Queue<Integer> q = new LinkedList<>();
        boolean[] visited = new boolean[arr.length];
        visited[0] = true;
        q.offer(0);
        while (!q.isEmpty()) {
            int size = q.size();
            for (int i = 0; i < size; i++) {
                int cur = q.poll();
                for (int j = inverse.get(arr[cur]).size() - 1; j >= 0; j--) {
                    int next = inverse.get(arr[cur]).get(j);
                    if (next != cur && !visited[next]) {
                        if (next == arr.length - 1) return steps + 1;
                        q.offer(next);
                        visited[next] = true;
                    }
                }
                if (cur > 0 && !visited[cur - 1]) {
                    q.offer(cur - 1);
                    visited[cur - 1] = true;
                }
                if (cur < arr.length - 1 && !visited[cur + 1]) {
                    if (cur + 1 == arr.length - 1) return steps + 1;
                    q.offer(cur + 1);
                    visited[cur + 1] = true;
                }
            }
            steps++;
        }
        return -1;
    }
}