题目描述

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]

题目思路

• 思路一、第 i 位保存 0~i 位的和，时间复杂度为 O(n)

class Solution {
public:
    vector<int> runningSum(vector<int>& nums) {
        for(int i = 1; i < nums.size(); ++i)
        {
            nums[i]+=nums[i-1];
        }
        return nums;
    }
};

总结展望