标签：constrain continue traints poll start NPU const dfs check
Given an array of non-negative integers arr, you are initially positioned at start index of the array. When you are at index i, you can jump to i + arr[i] or i - arr[i], check if you can reach to any index with value 0.
Notice that you can not jump outside of the array at any time.
Example 1:
Input: arr = [4,2,3,0,3,1,2], start = 5 Output: true Explanation: All possible ways to reach at index 3 with value 0 are: index 5 -> index 4 -> index 1 -> index 3 index 5 -> index 6 -> index 4 -> index 1 -> index 3
Example 2:
Input: arr = [4,2,3,0,3,1,2], start = 0 Output: true Explanation: One possible way to reach at index 3 with value 0 is: index 0 -> index 4 -> index 1 -> index 3
Example 3:
Input: arr = [3,0,2,1,2], start = 2 Output: false Explanation: There is no way to reach at index 1 with value 0.
Constraints:
1 <= arr.length <= 5 * 10^40 <= arr[i] < arr.length0 <= start < arr.lengthclass Solution { public static boolean canReach(int[] arr, int start) { int n = arr.length; HashSet<Integer> visited = new HashSet<>(); // visited set Queue<Integer> q = new LinkedList<>(); q.add(start); while (!q.isEmpty()) { int i = q.poll(); if (arr[i] == 0) return true; // found then return it if (visited.contains(i)) continue; // already visited than continue visited.add(i); if (i + arr[i] < n) q.add(i + arr[i]); if (i - arr[i] >= 0) q.add(i - arr[i]); } return false;// not found } }
用queue，很巧妙
class Solution { public boolean canReach(int[] arr, int start) { if(start<0 || start>=arr.length || arr[start]<0) return false; if(arr[start] == 0) return true; arr[start] = -arr[start]; return canReach(arr, start + arr[start]) || canReach(arr, start - arr[start]); } }
DFS也可
https://leetcode.com/problems/jump-game-iii/discuss/473221/Simple-Java-DFS-solution
class Solution { public boolean canReach(int[] arr, int start) { return travel(arr, start, new boolean[arr.length]); } private boolean travel(int[] arr, int index, boolean[] visited) { //Out of bounds check, found solution already, or visited, get out if(index < 0 || index >= arr.length || visited[index]) return false; visited[index]=true; if(arr[index] == 0) return true; return travel(arr, index + arr[index], visited) || travel(arr, index - arr[index], visited); } }
https://leetcode.com/problems/jump-game-iii/discuss/482295/Java-Simple-DFS-0(N)-Space-O(N)-time
标签：constrain continue traints poll start NPU const dfs check
原文地址：https://www.cnblogs.com/wentiliangkaihua/p/12205168.html
