Description#
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 <= 100nums.length % 2 == 01 <= nums[i] <= 100
Solutions#
Solution 1#
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| 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
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| 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;
}
}
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| 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;
}
};
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| 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
}
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| 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;
}
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| impl Solution {
pub fn decompress_rl_elist(nums: Vec<i32>) -> Vec<i32> {
let n = nums.len() >> 1;
let mut ans = Vec::new();
for i in 0..n {
for _ in 0..nums[2 * i] {
ans.push(nums[2 * i + 1]);
}
}
ans
}
}
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| /**
* Note: The returned array must be malloced, assume caller calls free().
*/
int* decompressRLElist(int* nums, int numsSize, int* returnSize) {
int size = 0;
for (int i = 0; i < numsSize; i += 2) {
size += nums[i];
}
int* ans = malloc(size * sizeof(int));
for (int i = 0, j = 0; j < numsSize; j += 2) {
for (int k = 0; k < nums[j]; k++) {
ans[i++] = nums[j + 1];
}
}
*returnSize = size;
return ans;
}
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