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Formatted question description: https://leetcode.ca/all/1313.html

1313. Decompress Run-Length Encoded List (Easy)

We are given a list nums of integers representing a list compressed with run-length encoding.

Consider each adjacent pair of elements [a, b] = [nums[2*i], nums[2*i+1]] (with i >= 0).  For each such pair, there are a elements with value b in 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] + [4,4,4,4] is [2,4,4,4].

 

Constraints:

  • 2 <= nums.length <= 100
  • nums.length % 2 == 0
  • 1 <= nums[i] <= 100

Related Topics:
Array

Similar Questions:

Solution 1.

  • class Solution {
    public int[] decompressRLElist(int[] nums) {
        List<Integer> list = new ArrayList<Integer>();
        int length = nums.length;
        for (int i = 0; i < length; i += 2) {
            int count = nums[i];
            int value = nums[i + 1];
            for (int j = 0; j < count; j++)
                list.add(value);
        }
        int size = list.size();
        int[] decompressed = new int[size];
        for (int i = 0; i < size; i++)
            decompressed[i] = list.get(i);
        return decompressed;
    }
}

############

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;
    }
}

  • // OJ: https://leetcode.com/problems/decompress-run-length-encoded-list/
// Time: O(N) where N is the length of the output array.
// Space: O(1)
class Solution {
public:
    vector<int> decompressRLElist(vector<int>& nums) {
        vector<int> ans;
        for (int i = 0; i < nums.size(); i += 2) {
            for (int j = 0; j < nums[i]; ++j) ans.push_back(nums[i + 1]);
        }
        return ans;
    }
};

  • 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



  • 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
}

  • 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;
}


  • 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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