[leetcode] 303. Range Sum Query - Immutable @ python-优快云博客

本文链接：https://blog.youkuaiyun.com/danspace1/article/details/88666311

博客围绕给定整数数组，求指定索引区间元素和的问题展开。给出两种解法，一是使用list slicing，二是构造列表存储累积和，通过累积和相减得到区间元素和，还给出了示例及相关数据。

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原题

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.
Accepted
129,573
Submissions
351,139

解法1

使用list slicing.

代码

class NumArray(object):

    def __init__(self, nums):
        """
        :type nums: List[int]
        """
        self.data = nums

    def sumRange(self, i, j):
        """
        :type i: int
        :type j: int
        :rtype: int
        """
        return sum(self.data[i:j+1])


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

解法2

构造列表, 储存累积的和, 初始化列表为[0], 遍历nums, 每次加累积的和加入列表, 然后返回self.data[j+1] - self.data[i] 即可

代码

class NumArray(object):

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

    def sumRange(self, i, j):
        """
        :type i: int
        :type j: int
        :rtype: int
        """
        return self.data[j+1]-self.data[i]


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