Problem
Given an integer array nums
, handle multiple queries of the following type:
- Calculate the sum of the elements of
nums
between indicesleft
andright
inclusive whereleft <= right
.
Implement the NumArray
class:
NumArray(int[] nums)
Initializes the object with the integer arraynums
.int sumRange(int left, int right)
Returns the sum of the elements ofnums
between indicesleft
andright
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 <= 10^4
-10^5 <= nums[i] <= 10^5
0 <= left <= right < nums.length
At most
10^4
calls will be made tosumRange
.
Solution
class NumArray {
private int[] sums;
public NumArray(int[] nums) {
sums = new int[nums.length];
for (int i = 0; i < nums.length; i++) {
if (i == 0) {
sums[i] = nums[i];
} else {
sums[i] = sums[i - 1] + nums[i];
}
}
}
public int sumRange(int i, int j) {
if (i == 0) {
return sums[j];
}
return sums[j] - sums[i - 1];
}
}
/**
* Your NumArray object will be instantiated and called as such:
* NumArray obj = new NumArray(nums);
* int param_1 = obj.sumRange(left,right);
*/
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).