【leetcode】1524. Number of Sub-arrays With Odd Sum

题目如下：

Given an array of integers arr. Return the number of sub-arrays with odd sum.

As the answer may grow large, the answer must be computed modulo 10^9 + 7. 

Example 1:

Input: arr = [1,3,5]
Output: 4
Explanation: All sub-arrays are [[1],[1,3],[1,3,5],[3],[3,5],[5]]
All sub-arrays sum are [1,4,9,3,8,5].
Odd sums are [1,9,3,5] so the answer is 4.

Example 2:

Input: arr = [2,4,6]
Output: 0
Explanation: All sub-arrays are [[2],[2,4],[2,4,6],[4],[4,6],[6]]
All sub-arrays sum are [2,6,12,4,10,6].
All sub-arrays have even sum and the answer is 0.

Example 3:

Input: arr = [1,2,3,4,5,6,7]
Output: 16

Example 4:

Input: arr = [100,100,99,99]
Output: 4

Example 5:

Input: arr = [7]
Output: 1

Constraints:

• 1 <= arr.length <= 10^5
• 1 <= arr[i] <= 100

解题思路：假设dp_odd[i] = v表示以第i个元素作为子数组的最后一个元素时，arr有v个和为奇数的子数组，dp_even[i] = v2 则表示，arr中有v2个和为偶数子数组。对于任意一个arr[i]，如果值为奇数，那么可以和前面和为偶数的子数组形成和为奇数的子数组，一共有 dp_odd[i] = dp_even[i-1] + 1 （这里加1是因为arr[i]可以独自组成一个和为奇数的子数组），而dp_even[i] = dp_odd[i-1]；同理arr[i]为偶数时原理是一样的。

代码如下：

class Solution(object):
    def numOfSubarrays(self, arr):
        """
        :type arr: List[int]
        :rtype: int
        """
        dp_odd = [0] * len(arr)
        dp_even = [0] * len(arr)
        if arr[0] % 2 == 0:
            dp_even[0] = 1
        else:dp_odd[0] = 1

        for i in range(1,len(arr)):
            if arr[i] % 2 == 0:
                dp_even[i] += (dp_even[i-1] + 1)
                dp_odd[i] += dp_odd[i-1]
            else:
                dp_even[i] += dp_odd[i-1]
                dp_odd[i] += (dp_even[i - 1] + 1)
        #print dp_even
        #print dp_odd
        return sum(dp_odd) % (10**9+7)