Leetcode

Merge Intervals

  • Time:O(n\log n)
  • Space:O(n)

C++

class Solution {
 public:
  vector<vector<int>> merge(vector<vector<int>>& intervals) {
    vector<vector<int>> ans;

    sort(begin(intervals), end(intervals));

    for (const auto& interval : intervals)
      if (ans.empty() || ans.back()[1] < interval[0])
        ans.push_back(interval);
      else
        ans.back()[1] = max(ans.back()[1], interval[1]);

    return ans;
  }
};

JAVA

class Solution {
  public int[][] merge(int[][] intervals) {
    List<int[]> ans = new ArrayList<>();

    Arrays.sort(intervals, (a, b) -> (a[0] - b[0]));

    for (int[] interval : intervals)
      if (ans.isEmpty() || ans.get(ans.size() - 1)[1] < interval[0])
        ans.add(interval);
      else
        ans.get(ans.size() - 1)[1] = Math.max(ans.get(ans.size() - 1)[1], interval[1]);

    return ans.toArray(new int[ans.size()][]);
  }
}

Python

class Solution:
  def merge(self, intervals: List[List[int]]) -> List[List[int]]:
    ans = []

    for interval in sorted(intervals):
      if not ans or ans[-1][1] < interval[0]:
        ans.append(interval)
      else:
        ans[-1][1] = max(ans[-1][1], interval[1])

    return ans