Jump Game III

Description

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

Code

我原本自己寫的 DFS，需要用 $O(n)$ 的 visited 。

class Solution {
public:
    bool canReach(vector<int>& arr, int start) {
        int n = arr.size();
        vector<int> visited(n, 0);
        return dfs(start, arr, visited);
    }
 
    bool dfs(int index, vector<int>& arr, vector<int>& visited) {
        if(visited[index]) return false;
        visited[index] = 1;
        if(arr[index] == 0) {
            return true;
        }
        bool left = false;
        bool right = false;
        if(index - arr[index] >= 0) left = dfs(index - arr[index], arr, visited);
        if(index + arr[index] < arr.size()) right = dfs(index + arr[index], arr, visited);
        return left || right;
    }
};

lee125 大神的解答，將 traverse 過的 element 的值設為負值，就不需要 visited 來檢查，且要找的值 0 就可以由 !(A[i] = -A[i] 檢查出來。

看完只能說，人的智商還是有差。

class Solution {
public:
    bool canReach(vector<int>& A, int i) {
        return 0 <= i && i < A.size() && A[i] >= 0 && (!(A[i] = -A[i]) || canReach(A, i + A[i]) || canReach(A, i - A[i]));
    }
};

Source

• Jump Game III - LeetCode