303. Range Sum Query - Immutable -Medium

Question

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.

Note:
1. You may assume that the array does not change.
2. There are many calls to sumRange function.

给出一个整数数组，找出在i到j之间的元素总和（i<=j，包含i和j）。数组不会改变，并且会多次调用sumRange函数

Example

Given nums = [-2, 0, 3, -5, 2, -1]

sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3

Solution

• 动态规划解。既然会多次调用，只需要记下前i个元素的总和即可。定义dp[i]：前i个元素的总和。递推式：dp[i] = dp[i - 1] + nums[i]，在计算i到j的元素总和只需要将两个相减即可 dp[j] - dp[i] + nums[i] (这里考虑i为0的情况，所以不减去dp[i - 1])

  class NumArray(object):
      def __init__(self, nums):
          """
          :type nums: List[int]
          """
          self.nums = nums
          self.dp = [0] * len(nums)
          for index in range(len(nums)):
              if index == 0:
                  self.dp[index] = nums[index]
              else:
                  self.dp[index] = nums[index] + self.dp[index - 1]
  
      def sumRange(self, i, j):
          """
          :type i: int
          :type j: int
          :rtype: int
          """
          return self.dp[j] - self.dp[i - 1] + self.nums[i]