303 Range Sum Query - Immutable

303. Range Sum Query - Immutable

1. Question

Given an integer arraynums, find the sum of the elements between indicesiandj(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:

1. You may assume that the array does not change.

2. There are many calls to sumRange function.

2. Implementation

(1) DP

class NumArray {
    int prefixSum[];
    public NumArray(int[] nums) {
        int n = nums.length;
        prefixSum = new int[n + 1];

        for (int i = 1; i <= n; i++) {
            prefixSum[i] = prefixSum[i - 1] + nums[i - 1];
        }
    }

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

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

3. Time & Space Complexity

DP: 时间复杂度: NumArray(): O(n), sumRange(): O(1), 空间复杂度O(n)