LeetCode 829. Consecutive Numbers Sum--B站筆試題--C++解法

LeetCode 829. Consecutive Numbers Sum–B站筆試題–C++解法

---

LeetCode題解專欄：LeetCode題解
我做的所有的LeetCode的題目都放在這個專欄裏，大部分題目Java和Python的解法都有。

---

題目地址：Consecutive Numbers Sum - LeetCode

---

Given a positive integer N, how many ways can we write it as a sum of consecutive positive integers?

Example 1:

Input: 5
Output: 2
Explanation: 5 = 5 = 2 + 3

Example 2:

Input: 9
Output: 3
Explanation: 9 = 9 = 4 + 5 = 2 + 3 + 4

Example 3:

Input: 15
Output: 4
Explanation: 15 = 15 = 8 + 7 = 4 + 5 + 6 = 1 + 2 + 3 + 4 + 5

Note: 1 <= N <= 10 ^ 9.

---

這道題目是2019-9-10的B站筆試原題，可惜我沒有事先看到，聽別人在牛客網上說才知道的。

這道題目最容易想到的解法就是暴力窮舉，時間複雜度O(N^2)，肯定超時。
使用等差數列，可以快一些：
C++解法如下：

class Solution {
public:
    int consecutiveNumbersSum(int N) {
        int result = 1;
        int a = N;
        int temp;
        for (int i = 1; i < a / 2 + 1; i++) {
            for (int j = i + 1; j <= a - i; j++) {
                temp = (i + j) * (j - i + 1) / 2;
                if (temp == a) {
                    result += 1;
                    break;
                } else if (temp > a) {
                    break;
                }
            }
        }
        return result;
    }
};

但上面的做法會超時，換一個思路，首項是x，有m項，那麼[x + (x + (m-1))]m / 2->x*m+m*(m-1)/2=n，只要遍歷m從1到sqrt(n)即可。

C++解法如下：

class Solution {
public:
    int consecutiveNumbersSum(int N) {
        int ans = 0;
        for (int m = 1;; m++) {
            int m_x = N - m * (m - 1) / 2;
            if (m_x <= 0)
                break;
            if (m_x % m == 0)
                ans++;
        }
        return ans;
    }
};

更快的C++解法如下：

class Solution {
public:
    int consecutiveNumbersSum(int N) {
        int sum = 0, ans = 0;
        for(int i = 1; sum < N; i++) {
            sum += i;
            if (((N-sum) % i) == 0)
                ans++;
        }
        return ans;
    }
};

思路是：
let N = k + (k+1) + (k+2) + (k+3) + … + (k+i-1) = ik + (1+2+3+… + i-1)
let sum(i) = (1+2+3+…+i-1), then we have
N = sum(i) + ik ==>i*k = N - sum(i)
Which means: for each i, we can calculate N-sum(i). If N-sum(i) can be divided by i, there is an answer with length i.