LeetCode 1269.Number of Ways to Stay in the Same Place After Some Steps
一、原题描述
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 <= 5001 <= arrLen <= 10^6
二、简要翻译
在一个长度为 arrLen的数组上，起始位置为0，每次的移动步骤可以选择往左，往右或者原地不动。问 在steps 次后，仍然位于起始位置的移动方式有多少种。对结果取模，mod = 10^9 + 7.
三、代码思路分析
- 动态规划的典型题目。
- 难度应该分到median更合适。
四、代码
1 public int numWays(int steps, int arrLen) {
2 int mod = (int) 1e9 + 7;
3 long[] dpPrev = new long[arrLen + 1];
4 long[] dp = new long[arrLen + 1];
5 long[] temp;
6 dpPrev[0] = 1;
7 for (int i = 1; i <= steps; i++) {
8 int max = Math.min(i, arrLen - 1);
9 dp[0] = (dpPrev[0] + dpPrev[1]) % mod;
10 for (int j = 1; j <= max; j++) {
11 dp[j] = (dpPrev[j - 1] + dpPrev[j] + dpPrev[j + 1]) % mod;
12 }
13 temp = dpPrev;
14 dpPrev = dp;
15 dp = temp;
16 }
17 return (int) dpPrev[0];
18 }
