動機
複習prefix sum
Problem
Given an integer array nums
, handle multiple queries of the following type:
- Calculate the sum of the elements of
nums
between indicesleft
andright
inclusive whereleft <= right
.
Implement the NumArray
class:
NumArray(int[] nums)
Initializes the object with the integer arraynums
.int sumRange(int left, int right)
Returns the sum of the elements ofnums
between indicesleft
andright
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 tosumRange
.
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]