# 303.Range Sum Query Immutable

**303.Range Sum Query Immutable**

难度:Easy

> 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。

```
示例：

给定 nums = [-2, 0, 3, -5, 2, -1]，求和函数为 sumRange()

sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3
说明:

你可以假设数组不可变。
会多次调用 sumRange 方法。
```

简单题：

```
class NumArray {
private:
    vector<int> num;
public:
    NumArray(vector<int>& nums) {
        num=nums;
    }
    
    int sumRange(int i, int j) {
        return accumulate(num.begin()+i, num.begin()+j+1,0);
    }
};

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


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