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:
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| 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].
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Example 2:
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| 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].
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Example 3:
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| Input: nums = [3,1,2,10,1]
Output: [3,4,6,16,17]
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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
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| 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;
}
};
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Summary
这道题目的分享到这里，感谢你的支持！
