LeetCode -- Range Sum Query - Immutable

题目：

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.

Given nums = [-2, 0, 3, -5, 2, -1]

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

Note:

You may assume that the array does not change.

There are many calls to sumRange function.

思路：

这道题是简单的动态规划的思想，参考了别人的思路后写出了

O(n)

的构建时间复杂度，

O(1)

的查询时间复杂度。构建一个

vector

记录

nums

的和，如下：

i	0	1	2	3	4	5	6
nums	-2	0	3	-5	2	-1
sum	0	-2	-2	1	-4	-2	-3

此时，

sumRange(i, j) = sum[j+1]-sum[i]

，

j+1

是因为为了使代码简洁，刚开始的时候向

sum

push了一个0。

C++代码实现：

class NumArray {
public:
NumArray(vector<int> nums) {
sum.push_back(0);
for(int i=0; i<nums.size(); ++i){
sum.push_back(sum.back() + nums[i]);
}
}

int sumRange(int i, int j) {
return sum[j+1]-sum[i];
}
private:
vector<int> sum;
};

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