303. Range Sum Query - Immutable

Easy

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:

  1. You may assume that the array does not change.
  2. There are many calls to sumRange function.
package leetcode.easy;

class NumArray1 {
private int[] data; public NumArray1(int[] nums) {
data = nums;
} public int sumRange(int i, int j) {
int sum = 0;
for (int k = i; k <= j; k++) {
sum += data[k];
}
return sum;
}
} class NumArray3 {
private int[] sum; public NumArray3(int[] nums) {
sum = new int[nums.length + 1];
for (int i = 0; i < nums.length; i++) {
sum[i + 1] = sum[i] + nums[i];
}
} public int sumRange(int i, int j) {
return sum[j + 1] - sum[i];
}
} /**
* Your NumArray object will be instantiated and called as such: NumArray obj =
* new NumArray(nums); int param_1 = obj.sumRange(i,j);
*/
public class RangeSumQueryImmutable {
@org.junit.Test
public void test1() {
int[] nums = { -2, 0, 3, -5, 2, -1 };
NumArray1 obj = new NumArray1(nums);
int param_1 = obj.sumRange(0, 2);
int param_2 = obj.sumRange(2, 5);
int param_3 = obj.sumRange(0, 5);
System.out.println(param_1);
System.out.println(param_2);
System.out.println(param_3);
} @org.junit.Test
public void test3() {
int[] nums = { -2, 0, 3, -5, 2, -1 };
NumArray3 obj = new NumArray3(nums);
int param_1 = obj.sumRange(0, 2);
int param_2 = obj.sumRange(2, 5);
int param_3 = obj.sumRange(0, 5);
System.out.println(param_1);
System.out.println(param_2);
System.out.println(param_3);
}
}

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