跳跃游戏 IV

/*
 * 思路：同样是使用广度优先搜索（BFS），但这次利用Java Stream API进行部分简化。
 * 首先用流来构建值到索引的映射。
 * 然后使用一个队列进行BFS搜索，记录访问过的节点并计算步数。
 */

import java.util.*;
import java.util.stream.Collectors;
import java.util.stream.IntStream;

public class Solution {
    public int minJumps(int[] arr) {
        if (arr.length <= 1) return 0;

        Map<Integer, List<Integer>> valueToIndices = IntStream.range(0, arr.length)
            .boxed()
            .collect(Collectors.groupingBy(i -> arr[i]));

        Queue<Integer> queue = new LinkedList<>();
        queue.offer(0);
        boolean[] visited = new boolean[arr.length];
        visited[0] = true;
        int steps = 0;

        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                int currentIndex = queue.poll();
                if (currentIndex == arr.length - 1) return steps;

                List<Integer> nextIndices = new ArrayList<>(valueToIndices.getOrDefault(arr[currentIndex], new ArrayList<>()));
                if (currentIndex + 1 < arr.length) nextIndices.add(currentIndex + 1);
                if (currentIndex - 1 >= 0) nextIndices.add(currentIndex - 1);

                for (int nextIndex : nextIndices) {
                    if (nextIndex >= 0 && nextIndex < arr.length && !visited[nextIndex]) {
                        queue.offer(nextIndex);
                        visited[nextIndex] = true;
                    }
                }
                nextIndices.clear();
            }
            steps++;
        }
        return -1;
    }
}