動機

複習prefix sum

Problem

Given an integer array nums, handle multiple queries of the following type:

  1. Calculate the sum of the elements of nums between indices left and right inclusive where left <= right.

Implement the NumArray class:

  • NumArray(int[] nums) Initializes the object with the integer array nums.
  • int sumRange(int left, int right) Returns the sum of the elements of nums between indices left and right inclusive (i.e. nums[left] + nums[left + 1] + ... + nums[right]).

 

Example 1:

Input[NumArray, sumRange, sumRange, sumRange][[[-2, 0, 3, -5, 2, -1]], [0, 2], [2, 5], [0, 5]]Output[null, 1, -1, -3]ExplanationNumArray numArray = new NumArray([-2, 0, 3, -5, 2, -1]);numArray.sumRange(0, 2); // return (-2) + 0 + 3 = 1numArray.sumRange(2, 5); // return 3 + (-5) + 2 + (-1) = -1numArray.sumRange(0, 5); // return (-2) + 0 + 3 + (-5) + 2 + (-1) = -3

 

Constraints:

  • 1 <= nums.length <= 104
  • -105 <= nums[i] <= 105
  • 0 <= left <= right < nums.length
  • At most 104 calls will be made to sumRange.

Sol

class NumArray:

    def __init__(self, nums: List[int]):
        tmp = [0]
        for n in nums:
            tmp.append(n+tmp[-1])
        self.sums = tmp

    def sumRange(self, left: int, right: int) -> int:
        return self.sums[right+1]-self.sums[left]