leetcode-303-easy

Range Sum Query - Immutable

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

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:

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
Constraints:

1 <= nums.length <= 104
-105 <= nums[i] <= 105
0 <= left <= right < nums.length
At most 104 calls will be made to sumRange.

思路一：直接计算

class NumArray {
    private int[] nums;

    public NumArray(int[] nums) {
        this.nums = nums;
    }

    public int sumRange(int left, int right) {
        int sum = 0;
        for (int i = left; i <= right; i++) {
            sum += nums[i];
        }

        return sum;
    }
}

思路二：看了一下官方题解，sumRange() 方法要被调用多次，所以可以提前计算数组 sum 值，减少重复计算

        sums = new int[n + 1];
        for (int i = 0; i < n; i++) {
            sums[i + 1] = sums[i] + nums[i];
        }