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l203 range sum query immutable

admin 11月 13, 2020 0

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.

解题思路

针对每次输入都计算直接使用sum 无需额外空间
计算sum值，根据index输入输出 需要空间

Python实现1

class (object):

    def __init__(self, nums):
        """
        :type nums: List[int]
        """
        self.nums = nums

    def sumRange(self, i, j):
        """
        :type i: int
        :type j: int
        :rtype: int
        """
        return sum(self.nums[i:j+1])

Runtime: 959 ms

Python实现2

class (object):

    def __init__(self, nums):
        """
        :type nums: List[int]
        """
        self.sums =[]
        s = 0
        for n in nums:
            s+=n
            self.sums.append(s)
        print self.sums
    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]

Runtime: 109 ms

Python实现3

class (object):

    def __init__(self, nums):
        """
        :type nums: List[int]
        """
        self.sums = [0]*(len(nums)+1)
        for i in xrange(1,len(nums)+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]

Runtime: 62ms