Range Sum Query - Immutable

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303.Range Sum Query - Immutable

Given an integer array nums, handle multiple queries of the following type:

  1. Calculate the sum of the elements of nums between indices left and right inclusive where left <= right.

Implement the NumArray class:

  • NumArray(int[] nums) Initializes the object with the integer array nums.
  • int sumRange(int left, int right) Returns the sum of the elements of nums between indices left and right inclusive (i.e. nums[left] + nums[left + 1] + ... + nums[right]).

Example 1:

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Input
["NumArray", "sumRange", "sumRange", "sumRange"]
[[[-2, 0, 3, -5, 2, -1]], [0, 2], [2, 5], [0, 5]]
Output
[null, 1, -1, -3]

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

时间复杂度：O(n2)

空间复杂度：O(n)

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class NumArray {
public:
    vector<int> arr;
    NumArray(vector<int>& nums) {
        arr = nums;
    }

    int sumRange(int left, int right) {
        int sum = 0;
        while(left<=right) {
            sum += arr[left];
            left++;
        }
        return sum;
    }
};

前缀和：

时间复杂度：O(n)

空间复杂度：O(n)

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class NumArray {
public:
    vector<int> sums;

    NumArray(vector<int>& nums) {
        int n = nums.size();
        sums.resize(n + 1);
        for (int i = 0; i < n; i++) {
            sums[i + 1] = sums[i] + nums[i];
        }
    }

    int sumRange(int i, int j) {
        return sums[j + 1] - sums[i];
    }
};