作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.cn/
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
保存累积和
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)
如果多用一个元素放在开头，那么上面的这个代码可以简化。
class NumArray(object):
def __init__(self, nums):
"""
:type nums: List[int]
"""
N = len(nums)
self.sums = [0] * (N + 1)
for i in range(1, N + 1):
self.sums[i] = self.sums[i - 1] + nums[i - 1]
def sumRange(self, i, j):
"""
:type i: int
:type j: int
:rtype: int
"""
return self.sums[j + 1] - self.sums[i]
# 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
2018 年 11 月 24 日 —— 周六快乐