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.
题目大意:
给定整数数组nums,计算下标i与j之间的元素和(i ≤ j),包含边界。
测试样例见题目描述。
注意:
- 你可以假设数组不会改变。
- sumRange函数会调用多次。
解题思路:
计算辅助数组sums:
sums[0] = 0 sums[i+1] = sums[i] + nums[i]
则sumRange(i, j) = sums[j+1] - sums[i]
Python代码:
class NumArray(object):
def __init__(self, nums):
"""
initialize your data structure here.
:type nums: List[int]
"""
size = len(nums)
self.sums = [0] * (size + 1)
for x in range(size):
self.sums[x + 1] += self.sums[x] + nums[x]
def sumRange(self, i, j):
"""
sum of elements nums[i..j], inclusive.
: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:
# numArray = NumArray(nums)
# numArray.sumRange(0, 1)
# numArray.sumRange(1, 2)
本文链接:http://bookshadow.com/weblog/2015/11/10/leetcode-range-sum-query-immutable/
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Joe 发布于 2015年11月10日 14:30 #
哈哈,你好快