---
# System prepended metadata

title: LeetCode 1313
tags: [python, LeetCode]

---

# LeetCode 1313
###### tags: `python`,`LeetCode`
>這邊使用Python解題
## 題目：
We are given a list nums of integers representing a list compressed with run-length encoding.

Consider each adjacent pair of elements [freq, val] = [nums[2*i], nums[2*i+1]] (with i >= 0).  For each such pair, there are freq elements with value val concatenated in a sublist. Concatenate all the sublists from left to right to generate the decompressed list.

Return the decompressed list.

## 範例
### 範例 1:
```
Input: nums = [1,2,3,4]
Output: [2,4,4,4]
Explanation: The first pair [1,2] means we have freq = 1 and val = 2 so we generate the array [2].
The second pair [3,4] means we have freq = 3 and val = 4 so we generate [4,4,4].
At the end the concatenation [2] + [4,4,4] is [2,4,4,4].
```
### 範例 2:
```
Input: nums = [1,1,2,3]
Output: [1,3,3]
```

## 條件限制
* 2 <= nums.length <= 100
* nums.length % 2 == 0
* 1 <= nums[i] <= 100

## 我的解題思路：
這題我一開始看到時，直覺就是會跟機偶數判斷，果不其然我的猜想被題目證實了XD
[freq, val] = [nums[2*i], nums[2*i+1]] (with i >= 0). 這段還蠻好理解的，所以我直接把這想法套到程式碼上：
## 程式碼:
```
def decompressRLElist(self, nums: List[int]) -> List[int]:
    result = []
    for i in range(len(nums)):
        if i % 2 == 0:
            for f in range(nums[i]):
                result.append(nums[i+1])
    return result
```