303. 区域和检索 - 数组不可变

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给定一个整数数组 nums,求出数组从索引 iji ≤ j)范围内元素的总和,包含 ij两点。

实现 NumArray 类:

  • NumArray(int[] nums) 使用数组 nums 初始化对象
  • int sumRange(int i, int j) 返回数组 nums 从索引 iji ≤ j)范围内元素的总和,包含 ij两点(也就是 sum(nums[i], nums[i + 1], ... , nums[j])

示例:

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输入:
["NumArray", "sumRange", "sumRange", "sumRange"]
[[[-2, 0, 3, -5, 2, -1]], [0, 2], [2, 5], [0, 5]]
输出:
[null, 1, -1, -3]

解释:
NumArray numArray = new NumArray([-2, 0, 3, -5, 2, -1]);
numArray.sumRange(0, 2); // return 1 ((-2) + 0 + 3)
numArray.sumRange(2, 5); // return -1 (3 + (-5) + 2 + (-1))
numArray.sumRange(0, 5); // return -3 ((-2) + 0 + 3 + (-5) + 2 + (-1))

提示:

  • 0 <= nums.length <= 104
  • -105 <= nums[i] <= 105
  • 0 <= i <= j < nums.length
  • 最多调用 104sumRange 方法

解答:

使用前缀和方法,前缀和表示的是[0,i)之间的和,比如对于数组 [1,2,3,4,5],presum[0]=0,presum[1]=0+1=1……

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for i in range(0,len(nums)+1):
presum[i+1]=presum[i]+nums[i]

因此本题解法如下:

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class NumArray:

def __init__(self, nums: List[int]):
_length=len(nums)
self.presum=[0]*(_length+1)
for i in range(0,_length):
self.presum[i+1]=self.presum[i]+nums[i]
def sumRange(self, i: int, j: int) -> int:
return self.presum[j+1]-self.presum[i]



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