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题目

303.区域和检索 - 数组不可变

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] <= 1050 <= i <= j < nums.length最多调用 104 次 sumRange 方法

题解(Java)

class NumArray {       private int[] nums;    public NumArray(int[] nums)     {           this.nums = nums;    }    public int sumRange(int i, int j)     {           int sum = 0;        for(int index = i;index <= j;index++)        {               sum += this.nums[index];        }        return sum;    }}/** * Your NumArray object will be instantiated and called as such: * NumArray obj = new NumArray(nums); * int param_1 = obj.sumRange(i,j); */