给定一个整数数组 nums
,求出数组从索引 i
到 j
(i ≤ j
)范围内元素的总和,包含 i
、j
两点。
实现 NumArray
类:
NumArray(int[] nums)
使用数组nums
初始化对象int sumRange(int i, int j)
返回数组nums
从索引i
到j
(i ≤ j
)范围内元素的总和,包含i
、j
两点(也就是sum(nums[i], nums[i + 1], ... , nums[j])
)
示例:
输入: ["NumArray", "sumRange", "sumRange", "sumRange"] [[[-2, 0, 3, -5, 2, -1]], [0, 2], [2, 5], [0, 5]] 输出: [null, 1, -1, -3] 解释: NumArray numArray = new NumArray([-2, 0, 3, -5, 2, -1]); numArray.sumRange(0, 2); // return 1 ((-2) + 0 + 3) numArray.sumRange(2, 5); // return -1 (3 + (-5) + 2 + (-1)) numArray.sumRange(0, 5); // return -3 ((-2) + 0 + 3 + (-5) + 2 + (-1))
提示:
0 <= nums.length <= 104
-105 <= nums[i] <= 105
0 <= i <= j < nums.length
- 最多调用
104
次sumRange
方法
C#代码
public class NumArray {
private int[] nums;
public NumArray(int[] nums) {
this.nums=nums;
}
public int SumRange(int i, int j) {
if(i==j)
return nums[i];
return nums[i]+SumRange(i+1,j);
}
}
/**
* Your NumArray object will be instantiated and called as such:
* NumArray obj = new NumArray(nums);
* int param_1 = obj.SumRange(i,j);
*/