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1345.Jump-Game-IV

1345. Jump Game IV

题目地址

题目描述

Given an array of integers arr, you are initially positioned at the first index of the array.
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In one step you can jump from index i to index:
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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.
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Notice that you can not jump outside of the array at any time.
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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.
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Example 2:
Input: arr = [7]
Output: 0
Explanation: Start index is the last index. You don't need to jump.
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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.
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Example 4:
Input: arr = [6,1,9]
Output: 2
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Example 5:
Input: arr = [11,22,7,7,7,7,7,7,7,22,13]
Output: 3
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Constraints:
1 <= arr.length <= 5 * 10^4
-10^8 <= arr[i] <= 10^8

代码

Approach #1 BFS

Time: O(N) && Space: O(N)
class Solution {
public int minJumps(int[] arr) {
int n = arr.length;
HashMap<Integer, List<Integer>> indicesOfValue = new HashMap<>(); // value => indices
for (int i = 0; i < n; i++)
indicesOfValue.computeIfAbsent(arr[i], x -> new LinkedList<>()).add(i);
boolean[] visited = new boolean[n];
Queue<Integer> q = new LinkedList<>();
q.offer(0);
visited[0] = true;
int step = 0;
while (!q.isEmpty()) {
for (int size = q.size(); size > 0; size--) {
int i = q.poll();
if (i == n - 1) return step; // reached to the last index
List<Integer> next = indicesOfValue.get(arr[i]);
next.add(i-1);
next.add(i+1);
for (int j : next) {
if (j >= 0 && j < n && !visited[j]) {
visited[j] = true;
q.offer(j);
}
}
}
step++;
}
return 0;
}
}