LeetCode 1306. Jump Game III

原题链接在这里：https://leetcode.com/problems/jump-game-iii/

题目：

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 * 104
• 0 <= arr[i] < arr.length
• 0 <= start < arr.length

题解：

Could use either BFS or DFS to check 0 could be reached.

Time Complexity: O(n). n = arr.length.

Space: O(n).

AC Java:

class Solution {
    public boolean canReach(int[] arr, int start) {
        int n = arr.length;
        boolean [] visited = new boolean[n];
        LinkedList<Integer> que = new LinkedList<>();
        que.add(start);
        visited[start] = true;
        
        while(!que.isEmpty()){
            int cur = que.poll();
            if(arr[cur] == 0){
                return true;
            }
            
            int can1 = cur - arr[cur];
            if(can1 >= 0 && can1 < n && !visited[can1]){
                que.add(can1);
                visited[can1] = true;
            }
            
            int can2 = cur + arr[cur];
            if(can2 >= 0 && can2 < n && !visited[can2]){
                que.add(can2);
                visited[can2] = true;
            }
        }
        
        return false;
    }
}

DFS version.

Time Complexity: O(n).

Space: O(n). stack space.

AC Java:

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]);
    }
}

类似Jump Game.