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] <= 10^6

思路

这里拿第三个例子来作介绍，Input:nums[3,1,2,10,1],Output:[3,4,6,16,17],我们假设获取的结果未get_nums[]

计算过程	结果
3	3
3 + 1	4
3+1+2 <=> 4+2	6
3+1+2+10 <=> 6+10	16
3+1+2+10+1 <=> 16+1	17

• 首先可是无论哪一个例子都有第一个值不变的情况，即get_nums[0] = nums[0]

• 后面的计算过程都可对应成
  get_nums[m] = get_nums[m-1] + nums[m]

• 由于计算过程在得出get_nums[m]后，不会再次使用nums[m]，我们则可直接取消get_nums[]使用nums[m]来避免内存的开销！

解法

java

class Solution {
    public int[] runningSum(int[] nums) {
        nums[0] = nums[0];

        for(int m = 1; m < nums.length;m++){
            nums[m] = nums[m] + nums[m-1];
        }
        return nums;
    }
}

python3

class Solution:
    def runningSum(self, nums: List[int]) -> List[int]: #这是新增的语法，为了说明参数和返回值的数据类型。不过仅仅的给人看的，实际上程序并不检查是否是相符的。
        nums[0] = nums[0]
        m = 1
        while m < len(nums):
            nums[m] = nums[m] + nums[m-1]
            m += 1
        return nums            

• sample 16ms submission

有点不太明白为啥这个运行较快?

class Solution:
    def runningSum(self, nums: List[int]) -> List[int]:
        var=[]
        for i in range(len(nums)):
            if i==0:
                var.append(nums[i])
            else:
                var.append(var[i-1]+nums[i])
        return var