303. Range Sum Query - Immutable

problem

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:
You may assume that the array does not change.
There are many calls to sumRange function.

solution

列表给定元素范围求和

因题目中‘There are many calls to sumRange function.’
直接每次请求，切片求和不满足效率要求

class NumArray(object):
    def __init__(self, nums):
        """
        initialize your data structure here.
        :type nums: List[int]
        """
        self.num = []
        a = 0
        for x in nums:
            a += x
            self.num.append(a)

    def sumRange(self, i, j):
        """
        sum of elements nums[i..j], inclusive.
        :type i: int
        :type j: int
        :rtype: int
        """
        
        if 0 == i:
            return self.num[j]
        else:
            return self.num[j]  - self.num[i-1]


# Your NumArray object will be instantiated and called as such:
# numArray = NumArray(nums)
# numArray.sumRange(0, 1)
# numArray.sumRange(1, 2)