【LeetCode】303. Range Sum Query - Immutable 解题报告

【LeetCode】303. Range Sum Query - Immutable 解题报告

标签（空格分隔）： LeetCode

题目地址：https://leetcode.com/problems/range-sum-query-immutable/description/

题目描述：

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.

解题方法

方法一：

可以直接用切片求和的方法做，也能A，但是效率太慢。

下面这个方式可以先把sums求出来，然后再调用的时候直接右边的sums-左边的sums即可得到结果。

class NumArray(object):

def __init__(self, nums):
"""
:type nums: List[int]
"""
self.sums = [0] * len(nums)
total = 0
for i, num in enumerate(nums):
total += num
self.sums[i] = total

def sumRange(self, i, j):
"""
:type i: int
:type j: int
:rtype: int
"""
if i == 0:
return self.sums[j]
else:
return self.sums[j] - self.sums[i - 1]

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

日期

2018 年 2 月 4 日