标签：mod pre not step turn time ace HERE bsp
You have a pointer at index 00 in an array of size arrLenarrLen. At each step, you can move 11 position to the left, 11 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 stepssteps and arrLenarrLen, return the number of ways such that your pointer still at index 00 after exactly stepssteps steps.
Since the answer may be too large, return it modulo 10^9 + 710?9??+7.
Example 1:
Input: 3 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
public class Solution { /** * @param steps: steps you can move * @param arrLen: the length of the array * @return: Number of Ways to Stay in the Same Place After Some Steps */ public int numWays(int steps, int arrLen) { // write your code here return helper(steps, arrLen, 0); } private int helper(int steps, int arrLen, int curPos) { if (curPos < 0 || curPos >= arrLen) { return 0; } if (steps == 0) { if (curPos == 0) { return 1; } return 0; } int lStep = helper(steps - 1, arrLen, curPos - 1); int rStep = helper(steps - 1, arrLen, curPos + 1); int sStep = helper(steps - 1, arrLen, curPos); return lStep + rStep + sStep; } }
[LintCode] 1835. Number of Ways to Stay in the Same Place After Some Steps I
标签：mod pre not step turn time ace HERE bsp
原文地址：https://www.cnblogs.com/xuanlu/p/12445988.html
