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943. Range Sum Query - Immutable
Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.
样例
Given nums = [-2, 0, 3, -5, 2, -1]
sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3
注意事项
- You may assume that the array does not change.
- There are many calls to
sumRangefunction.
解题思路：
因为会多次计算，所以我们提前把和都计算好，这样只用遍历一次数组。需要的时候直接用两个和相减就得到区间数字的和。
class NumArray(object):
def __init__(self, nums):
"""
:type nums: List[int]
"""
self.dp = [0] * len(nums)
num_sum = 0
for i in range(len(nums)):
num_sum += nums[i]
self.dp[i] = num_sum
def sumRange(self, i, j):
"""
:type i: int
:type j: int
:rtype: int
"""
if i == 0:
return self.dp[j]
else:
return self.dp[j] - self.dp[i-1]
# Your NumArray object will be instantiated and called as such:
# obj = NumArray(nums)
# param_1 = obj.sumRange(i,j)