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.
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
class Solution:
def decompressRLElist(self, nums: List[int]) -> List[int]:
res = []
for i in range(1, len(nums), 2):
res.extend([nums[i]] * nums[i - 1])
return res
class Solution {
public int[] decompressRLElist(int[] nums) {
int n = 0;
for (int i = 0; i < nums.length; i += 2) {
n += nums[i];
}
int[] res = new int[n];
for (int i = 1, k = 0; i < nums.length; i += 2) {
for (int j = 0; j < nums[i - 1]; ++j) {
res[k++] = nums[i];
}
}
return res;
}
}
function decompressRLElist(nums: number[]): number[] {
let n = nums.length >> 1;
let ans = [];
for (let i = 0; i < n; i++) {
let freq = nums[2 * i],
val = nums[2 * i + 1];
ans.push(...new Array(freq).fill(val));
}
return ans;
}
class Solution {
public:
vector<int> decompressRLElist(vector<int>& nums) {
vector<int> res;
for (int i = 1; i < nums.size(); i += 2) {
for (int j = 0; j < nums[i - 1]; ++j) {
res.push_back(nums[i]);
}
}
return res;
}
};
func decompressRLElist(nums []int) []int {
var res []int
for i := 1; i < len(nums); i += 2 {
for j := 0; j < nums[i-1]; j++ {
res = append(res, nums[i])
}
}
return res
}