[Leetcode] 303. Range Sum Query - Immutable 解题报告

题目：

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

Note:

You may assume that the array does not change.
There are many calls to sumRange function.
思路：

记录一个和数组sums，其中sums[j]表示从nums[0]到nums[j]的和。sums的状态转移方程是：sums[j] = sums[j - 1] + nums[j]，可以通过对数组一遍扫描完成；计算i到j之间的和可以通过sums[i, j] = sums[j] - sums[i-1]获得。

代码：

class NumArray {
public:
NumArray(vector<int> nums) { // the time complexity is O(n)
if (nums.size() > 0) {
sums.push_back(nums[0]);
for (int i = 1; i < nums.size(); ++i) {
sums.push_back(sums[i - 1] + nums[i]);
}
}
}

int sumRange(int i, int j) { // the time complexity is O(1)
if (i == 0) {
return sums[j];
}
else {
return sums[j] - sums[i - 1];
}
}

private:
vector<int> sums;
};

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