【leetcode】1269. Number of Ways to Stay in the Same Place After Some Steps

题目如下：

You have a pointer at index 0 in an array of size arrLen. At each step, you can move 1 position to the left, 1 position to the right in the array or stay in the same place  (The pointer should not be placed outside the array at any time).

Given two integers steps and arrLen, return the number of ways such that your pointer still at index 0 after exactly steps steps.

Since the answer may be too large, return it modulo 10^9 + 7.

Example 1:

Input: steps = 3, arrLen = 2
Output: 4
Explanation: There are 4 differents ways to stay at index 0 after 3 steps.
Right, Left, Stay
Stay, Right, Left
Right, Stay, Left
Stay, Stay, Stay

Example 2:

Input: steps = 2, arrLen = 4
Output: 2
Explanation: There are 2 differents ways to stay at index 0 after 2 steps
Right, Left
Stay, Stay

Example 3:

Input: steps = 4, arrLen = 2
Output: 8

Constraints:

• 1 <= steps <= 500
• 1 <= arrLen <= 10^6

解题思路：记dp[i][j]为移动i次后恰好位于下标j的次数，要使得第i步移动到j，那么第i-1步所处的位置就只能是 [j-1，j，j+1]，所以有dp[i][j] = dp[i-1][j-1] + dp[i-1][j] + dp[i-1][j+1] 。

代码如下：

class Solution(object):
    def numWays(self, steps, arrLen):
        """
        :type steps: int
        :type arrLen: int
        :rtype: int
        """
        arrLen = min(arrLen,steps)

        dp = [[0] * (arrLen) for _ in range(steps+1)]

        dp[1][0] = 1
        dp[1][1] = 1

        for i in range(1,steps+1):
            for j in range(len(dp[i])):
                dp[i][j] += dp[i-1][j]
                if j - 1 >= 0 and j - 1 < len(dp[i]):
                    dp[i][j] += dp[i-1][j-1]
                if j + 1 < len(dp[i]):
                    dp[i][j] += dp[i-1][j+1]

        return dp[-1][0] % (10**9 + 7)
        

原文地址：https://www.cnblogs.com/seyjs/p/12041899.html