Range Sum Query - Immutable

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.

Example:

Given nums = [-2, 0, 3, -5, 2, -1]

sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3

Note:

1. You may assume that the array does not change.

2. There are many calls to sumRange function.

题目说了数组不变，而且会quary很多次。那么明示用 prefix sum array 的方法来做

class NumArray(object):

    def __init__(self, nums):
        """
        :type nums: List[int]
        """
        length = len(nums)
        self.prefixsum = [0 for _ in xrange(length + 1)]
        for i, n in enumerate(nums, 1):
            self.prefixsum[i] = self.prefixsum[i - 1] + n   


    def sumRange(self, i, j):
        """
        :type i: int
        :type j: int
        :rtype: int
        """
        if i > j or j >= len(self.prefixsum) - 1:
            return 0
        return self.prefixsum[j + 1] - self.prefixsum[i]


# Your NumArray object will be instantiated and called as such:
# obj = NumArray(nums)
# param_1 = obj.sumRange(i,j)