LeetCode #1306 Jump Game III 跳跃游戏 III

1306 Jump Game III 跳跃游戏 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:

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 * 10^4
0 <= arr[i] < arr.length
0 <= start < arr.length

题目描述:

这里有一个非负整数数组 arr,你最开始位于该数组的起始下标 start 处。当你位于下标 i 处时,你可以跳到 i + arr[i] 或者 i - arr[i]。

请你判断自己是否能够跳到对应元素值为 0 的 任一 下标处。

注意,不管是什么情况下,你都无法跳到数组之外。

示例:

示例 1:

输入:arr = [4,2,3,0,3,1,2], start = 5
输出:true
解释:
到达值为 0 的下标 3 有以下可能方案:
下标 5 -> 下标 4 -> 下标 1 -> 下标 3
下标 5 -> 下标 6 -> 下标 4 -> 下标 1 -> 下标 3

示例 2:

输入:arr = [4,2,3,0,3,1,2], start = 0
输出:true
解释:
到达值为 0 的下标 3 有以下可能方案:
下标 0 -> 下标 4 -> 下标 1 -> 下标 3

示例 3:

输入:arr = [3,0,2,1,2], start = 2
输出:false
解释:无法到达值为 0 的下标 1 处。

提示:

1 <= arr.length <= 5 * 10^4
0 <= arr[i] < arr.length
0 <= start < arr.length

思路:

DFS
分别搜索两个方向直到超出边界或者到达已经搜索过的下标
可以原地修改数组
时间复杂度为 O(n), 空间复杂度为 O(1)

代码:

C++:

class Solution 
{
public:
    bool canReach(vector& arr, int start) 
    {
        return dfs(arr, start);
    }
private:
    bool dfs(vector& arr, int cur)
    {
        if (cur < 0 or cur >= arr.size() or arr[cur] == -1) return false;
        int step = arr[cur];
        arr[cur] = -1;
        return !step or dfs(arr, cur + step) or dfs(arr, cur - step);
    }
};

Java:

class Solution {
    public boolean canReach(int[] arr, int start) {
        return dfs(arr, start);
    }

    private boolean dfs(int[] arr, int cur) {
        if (cur < 0 || cur >= arr.length || arr[cur] == -1) return false;
        int step = arr[cur];
        arr[cur] = -1;
        return step == 0 || dfs(arr, cur + step) || dfs(arr, cur - step);
    }
}

Python:

class Solution:
    def canReach(self, arr: List[int], start: int) -> bool:
        def dfs(cur: int) -> bool:
            if not -1 < cur < len(arr) or arr[cur] == -1:
                return False
            step, arr[cur] = arr[cur], -1
            return not step or dfs(cur + step) or dfs(cur - step)
        return dfs(start)

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