Halo

A magic place for coding

0%

LeetCode 解题报告(364) -- 1480. Running Sum of 1d Array

Problem

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:

1
2
3
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:

1
2
3
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:

1
2
Input: nums = [3,1,2,10,1]
Output: [3,4,6,16,17]

Constraints:

  • 1 <= nums.length <= 1000
  • -10^6 <= nums [i] <= 10^6

Analysis

   这道题目就是一维数组简单的 presum 了,从头一直往后加就行。


Solution

   无。


Code

1
2
3
4
5
6
7
8
9
10
11
12
class Solution {
public:
vector<int> runningSum(vector<int>& nums) {
int size = nums.size ();
vector<int> result(size, 0);
result [0] = nums [0];
for (int i = 1; i < size; i++) {
result [i] = result [i - 1] + nums [i];
}
return result;
}
};

Summary

   这道题目的分享到这里,感谢你的支持!

Welcome to my other publishing channels