问题：
给定一个数组，相邻两两元素构成编码，第一个元素代表出现次数，第二个元素代表出现的元素。
求编码前的原数组。
Example 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]. Example 2: Input: nums = [1,1,2,3] Output: [1,3,3] Constraints: 2 <= nums.length <= 100 nums.length % 2 == 0 1 <= nums[i] <= 100
解法：
遍历编码数组，每次遍历取出一个编码对，两个元素 i+=2。
再循环出现次数freq次，将元素nums[i+1] push_back到结果中。
代码参考：
1 class Solution { 2 public: 3 vector<int> decompressRLElist(vector<int>& nums) { 4 vector<int> res; 5 for(int i=0; i+1<nums.size(); i+=2){ 6 for(int freq=nums[i];freq>0;freq--){ 7 res.push_back(nums[i+1]); 8 } 9 } 10 return res; 11 } 12 };
1313. Decompress Run-Length Encoded List
原文：https://www.cnblogs.com/habibah-chang/p/13234671.html