leetcode 303. Range Sum Query - Immutable

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

前缀和思想。如果修改的次数比较多，那么就可以用线段树了,

class NumArray {
public:
    vector<int> sum;
    NumArray(vector<int> nums) {
        for (int i = 0; i < nums.size(); ++i) {
            if (sum.size()) sum.push_back(nums[i] + sum.back());
            else sum.push_back(nums[i]);
        }
    }
    
    int sumRange(int i, int j) {
        if (i < 1) return sum[j];
        else return sum[j] - sum[i-1];
    }
};

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