How to Compute Running Sum of 1d Array using std::partial_sum in
- 时间:2020-09-08 11:19:41
- 分类:网络文摘
- 阅读:154 次
Given an array nums. We define a running sum of an array as runningSum[i] = sum(nums[0]…nums[i]). Return the running sum of nums.
Example 1:
Input: nums = [1,2,3,4]
Output: [1,3,6,10]
Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4].Example 2:
Input: nums = [1,1,1,1,1]
Output: [1,2,3,4,5]
Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1].Example 3:
Input: nums = [3,1,2,10,1]
Output: [3,4,6,16,17]Constraints:
1 <= nums.length <= 1000
-10^6 <= nums[i] <= 10^6Hints:
Think about how we can calculate the i-th number in the running sum from the (i-1)-th number.
Accumulate Prefix Sum
A traditional approach to compute the running sum would be to accumulate the prefix sum. You can use an additional variable to keep the sum, or simply we can update in place (use previous element in the array) to store the prefix sum.
1 2 3 4 5 6 7 8 9 | class Solution { public: vector<int> runningSum(vector<int>& nums) { for (int i = 1; i < nums.size(); ++ i) { nums[i] += nums[i - 1]; } return nums; } }; |
class Solution {
public:
vector<int> runningSum(vector<int>& nums) {
for (int i = 1; i < nums.size(); ++ i) {
nums[i] += nums[i - 1];
}
return nums;
}
};C++ std::partial_sum
The C++ std::partial_sum does exactly this job. The function computes the partial sums of the elements in the range specified by [first, last) and update them to the range begining at third parameter.
1 2 3 4 5 6 7 | class Solution { public: vector<int> runningSum(vector<int>& nums) { partial_sum(begin(nums), end(nums), begin(nums)); return nums; } }; |
class Solution {
public:
vector<int> runningSum(vector<int>& nums) {
partial_sum(begin(nums), end(nums), begin(nums));
return nums;
}
};The std::partial_sum may be implemented as follows:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 | template<class InputIt, class OutputIt> constexpr // since C++20 OutputIt partial_sum(InputIt first, InputIt last, OutputIt d_first) { if (first == last) return d_first; typename std::iterator_traits<inputit>::value_type sum = *first; *d_first = sum; while (++first != last) { sum = std::move(sum) + *first; // std::move since C++20 *++d_first = sum; } return ++d_first; // or, since C++14: // return std::partial_sum(first, last, d_first, std::plus<>()); } |
template<class InputIt, class OutputIt>
constexpr // since C++20
OutputIt partial_sum(InputIt first, InputIt last,
OutputIt d_first)
{
if (first == last) return d_first;
typename std::iterator_traits<inputit>::value_type sum = *first;
*d_first = sum;
while (++first != last) {
sum = std::move(sum) + *first; // std::move since C++20
*++d_first = sum;
}
return ++d_first;
// or, since C++14:
// return std::partial_sum(first, last, d_first, std::plus<>());
}The loops can be unrolled as the following:
1 2 3 4 5 | *(d_first) = *first; *(d_first+1) = *first + *(first+1); *(d_first+2) = *first + *(first+1) + *(first+2); *(d_first+3) = *first + *(first+1) + *(first+2) + *(first+3); ... |
*(d_first) = *first; *(d_first+1) = *first + *(first+1); *(d_first+2) = *first + *(first+1) + *(first+2); *(d_first+3) = *first + *(first+1) + *(first+2) + *(first+3); ...
–EOF (The Ultimate Computing & Technology Blog) —
推荐阅读:一个数倒过来看是正着看的1.5倍 牛吃草问题练习题 稍复杂的牛吃草问题 车站检票口的牛吃草问题 和自动扶梯有关的牛吃草问题 草每天匀速减少的牛吃草问题 让明月清风走进心灵 小小说 无题 那一道风景 坦然面对失败
- 评论列表
-
- 添加评论