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1480. Running Sum of 1d Array
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:
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1 <= nums.length <= 1000 -
-10^6 <= nums[i] <= 10^6
思路分析
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/**
* @author D瓜哥 · https://www.diguage.com
* @since 2024-09-13 23:14:43
*/
public int[] runningSum(int[] nums) {
int[] dp = new int[nums.length];
dp[0] = nums[0];
for (int i = 1; i < nums.length; i++) {
dp[i] = nums[i] + dp[i - 1];
}
return dp;
}