LeetCode 2400 Number of Ways to Reach a Position After Exactly k Steps (dp 或 组合数 推荐)

You are given two positive integers startPos and endPos. Initially, you are standing at position startPos on an infinite number line. With one step, you can move either one position to the left, or one position to the right.

Given a positive integer k, return the number of different ways to reach the position endPos starting from startPos, such that you perform exactly k steps. Since the answer may be very large, return it modulo 109 + 7.

Two ways are considered different if the order of the steps made is not exactly the same.

Note that the number line includes negative integers.

Example 1:

Input: startPos = 1, endPos = 2, k = 3
Output: 3
Explanation: We can reach position 2 from 1 in exactly 3 steps in three ways:
- 1 -> 2 -> 3 -> 2.
- 1 -> 2 -> 1 -> 2.
- 1 -> 0 -> 1 -> 2.
It can be proven that no other way is possible, so we return 3.

Example 2:

Input: startPos = 2, endPos = 5, k = 10
Output: 0
Explanation: It is impossible to reach position 5 from position 2 in exactly 10 steps.

Constraints:

• 1 <= startPos, endPos, k <= 1000

题目链接：https://leetcode.com/problems/number-of-ways-to-reach-a-position-after-exactly-k-steps/

题目大意：直线给两点，从起点到终点走k步的方案数

题目分析：最容易想到的是记忆化搜索，dp[i][j]表示从位置i到终点走k步的方案数

dp[i][j] = dp[abs(i - 1)][j - 1] + dp[i + 1][j - 1]

94ms，时间击败82.98%

class Solution {
    
    int[][] dp = new int[1001][1001];
    final int mod = 1000000007;
    
    int dfs(int d, int k) {
        if (d > k) {
            return 0;
        }
        if (d == k) {
            return 1;
        }
        if (dp[d][k] != 0) {
            return dp[d][k];
        }
        int ans = (dfs(Math.abs(d - 1), k - 1) + dfs(d + 1, k - 1)) % mod;
        dp[d][k] = ans;
        return ans;
    }
    
    public int numberOfWays(int startPos, int endPos, int k) {
        int d = Math.abs(endPos - startPos);
        if (d % 2 != k % 2 || d > k) {
            return 0;
        }
        return dfs(d, k);
    }
}

组合数方法：

设向右走x步，向左走y步，假设endPos比startPos大则

x - y = endPos - startPos

x + y = k

联立得x = (endPos - startPos + k) / 2，答案即为C(n, x)，取模需要用到除法逆元，由费马小定理a^(p - 1) ≡ 1 mod p

a * a^(p - 2) ≡ 1 mod p

b * a^(p - 2) ≡ b / a mod p

C(n, x) = n * (n - 1) * ... * (n - i + 1) / x! (0 < i <= x)

2ms，时间击败98.14%

class Solution {

    // x + y = k
    // y - x = s - e
    // x = (k + s - e) / 2

    final int mod = 1000000007;
    
    int qpow(long x, int n) {
        long ans = 1;
        while (n != 0) {
            if ((n & 1) == 1) {
                ans = (ans * x) % mod;
            }
            n >>= 1;
            x = (x * x) % mod;
        }
        return (int)ans;
    }

    int C(int n, int k) {
        if (n - k < k) {
            k = n - k;
        }
        long ans = 1;
        for (int i = 1; i <= k; i++) {
            long b = (long)((n - i + 1) % mod);
            long a = (long)(i % mod);
            ans = ans * (b * qpow(a, mod - 2) % mod) % mod;
        }
        return (int)ans;
    }
    
    public int numberOfWays(int startPos, int endPos, int k) {
        int d = Math.abs(endPos - startPos);
        if (d % 2 != k % 2 || d > k) {
            return 0;
        }
        int l = (k + d) >> 1;
        return C(k, l);
    }
}