【LEETCODE】303-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.

题意：

给一个整数数组，和角标 i and j (i ≤ j),

计算 i 到 j 的所有数的和，包括 i 和 j 位的数字

参考：

http://www.tuicool.com/articles/mIzeUrU

思路：

因为 nums 和 i，j 分开为两个函数的参数，调用方式又是 numArray.sumRange(1, 2)

所以要建立两个函数之间的联系

def __init__(self, nums) 中得到 从 nums[0] 到 nums[i－1] 的和 sum[i-1]

def sumRange(self, i, j) 得到 nums[i] 到 nums[j] 的和 sum[j]-sum[i-1]

 

class NumArray(object):
    def __init__(self, nums):
        """
        initialize your data structure here.
        :type nums: List[int]
        """
        
        #if nums is None:
            #return None
        
        self.sum=[0 for i in range(len(nums)+1)]   #记得加self AttributeError: 'NumArray' object has no attribute 'sum'
        #self.sum[0]=nums[0]                       #需要在nums[0]之前加一个sum＝0位，IndexError: list index out of range
        
        for i in range(len(nums)):
            self.sum[i+1]=self.sum[i]+nums[i]
        
        
    def sumRange(self, i, j):
        """
        sum of elements nums[i..j], inclusive.
        :type i: int
        :type j: int
        :rtype: int
        """
        return self.sum[j+1]-self.sum[i]
        


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