303. Range Sum Query - Immutable*

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本文介绍了一种高效的区间求和方法，通过预处理数组来快速计算任意两个索引之间的元素之和。这种方法适用于固定不变的数组，并且需要频繁进行区间求和操作的场景。

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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.

My code:

class NumArray(object):
    def __init__(self, nums):
        """
        initialize your data structure here.
        :type nums: List[int]
        """
        self.nums = nums
        self.sums = [0 for i in range(len(nums))]
        for index in range(len(nums)):
            if index == 0:
                self.sums[index] = self.nums[index]
            else:
                self.sums[index]= self.sums[index-1]+self.nums[index]        

    def sumRange(self, i, j):
        """
        sum of elements nums[i..j], inclusive.
        :type i: int
        :type j: int
        :rtype: int
        """
        if i<0 or j<0 or i>len(self.nums) or j>len(self.nums) or i>j:
            return 0
        elif i==0:
            return self.sums[j]
        else:
            return self.sums[j]-self.sums[i-1]

public class NumArray {
    int[] sums;

    public NumArray(int[] nums) {
        for(int i=1;i<nums.length;i++){
            nums[i]=nums[i-1]+nums[i];
        }
        this.sums=nums;
    }
    
    public int sumRange(int i, int j) {
        if(i==0) return this.sums[j];
        else  return this.sums[j]-this.sums[i-1]; 
        
    }
}

/**
 * Your NumArray object will be instantiated and called as such:
 * NumArray obj = new NumArray(nums);
 * int param_1 = obj.sumRange(i,j);
 */