Question

56. Merge Intervals

中文文档

Description

Given an array of intervals where intervals[i] = [starti, endi], merge all overlapping intervals, and return _an array of the non-overlapping intervals that cover all the intervals in the input_.

 

Example 1:

Input: intervals = [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlap, merge them into [1,6].

Example 2:

Input: intervals = [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considered overlapping.

 

Constraints:

• 1 <= intervals.length <= 104
• intervals[i].length == 2
• 0 <= starti <= endi <= 104

Solutions

Solution 1: Sorting + One-pass Traversal

We can sort the intervals in ascending order by the left endpoint, and then traverse the intervals for merging operations.

The specific merging operation is as follows.

First, we add the first interval to the answer. Then, we consider each subsequent interval in turn:

• If the right endpoint of the last interval in the answer array is less than the left endpoint of the current interval, it means that the two intervals will not overlap, so we can directly add the current interval to the end of the answer array;
• Otherwise, it means that the two intervals overlap. We need to use the right endpoint of the current interval to update the right endpoint of the last interval in the answer array, setting it to the larger of the two.

Finally, we return the answer array.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(\log n)$. Here, $n$ is the number of intervals.

class Solution:
  def merge(self, intervals: List[List[int]]) -> List[List[int]]:
    intervals.sort()
    ans = []
    st, ed = intervals[0]
    for s, e in intervals[1:]:
      if ed < s:
        ans.append([st, ed])
        st, ed = s, e
      else:
        ed = max(ed, e)
    ans.append([st, ed])
    return ans

class Solution {
  public int[][] merge(int[][] intervals) {
    Arrays.sort(intervals, Comparator.comparingInt(a -> a[0]));
    int st = intervals[0][0], ed = intervals[0][1];
    List<int[]> ans = new ArrayList<>();
    for (int i = 1; i < intervals.length; ++i) {
      int s = intervals[i][0], e = intervals[i][1];
      if (ed < s) {
        ans.add(new int[] {st, ed});
        st = s;
        ed = e;
      } else {
        ed = Math.max(ed, e);
      }
    }
    ans.add(new int[] {st, ed});
    return ans.toArray(new int[ans.size()][]);
  }
}

class Solution {
public:
  vector<vector<int>> merge(vector<vector<int>>& intervals) {
    sort(intervals.begin(), intervals.end());
    int st = intervals[0][0], ed = intervals[0][1];
    vector<vector<int>> ans;
    for (int i = 1; i < intervals.size(); ++i) {
      if (ed < intervals[i][0]) {
        ans.push_back({st, ed});
        st = intervals[i][0];
        ed = intervals[i][1];
      } else {
        ed = max(ed, intervals[i][1]);
      }
    }
    ans.push_back({st, ed});
    return ans;
  }
};

func merge(intervals [][]int) (ans [][]int) {
  sort.Slice(intervals, func(i, j int) bool {
    return intervals[i][0] < intervals[j][0]
  })
  st, ed := intervals[0][0], intervals[0][1]
  for _, e := range intervals[1:] {
    if ed < e[0] {
      ans = append(ans, []int{st, ed})
      st, ed = e[0], e[1]
    } else if ed < e[1] {
      ed = e[1]
    }
  }
  ans = append(ans, []int{st, ed})
  return ans
}

function merge(intervals: number[][]): number[][] {
  intervals.sort((a, b) => a[0] - b[0]);
  const ans: number[][] = [];
  let [st, ed] = intervals[0];
  for (const [s, e] of intervals.slice(1)) {
    if (ed < s) {
      ans.push([st, ed]);
      [st, ed] = [s, e];
    } else {
      ed = Math.max(ed, e);
    }
  }
  ans.push([st, ed]);
  return ans;
}

impl Solution {
  pub fn merge(mut intervals: Vec<Vec<i32>>) -> Vec<Vec<i32>> {
    intervals.sort_unstable_by(|a, b| a[0].cmp(&b[0]));
    let n = intervals.len();
    let mut res = vec![];
    let mut i = 0;
    while i < n {
      let l = intervals[i][0];
      let mut r = intervals[i][1];
      i += 1;
      while i < n && r >= intervals[i][0] {
        r = r.max(intervals[i][1]);
        i += 1;
      }
      res.push(vec![l, r]);
    }
    res
  }
}

public class Solution {
  public int[][] Merge(int[][] intervals) {
    intervals = intervals.OrderBy(a => a[0]).ToArray();
    int st = intervals[0][0], ed = intervals[0][1];
    var ans = new List<int[]>();
    for (int i = 1; i < intervals.Length; ++i) {
      if (ed < intervals[i][0]) {
        ans.Add(new int[] { st, ed });
        st = intervals[i][0];
        ed = intervals[i][1];
      } else {
        ed = Math.Max(ed, intervals[i][1]);
      }
    }
    ans.Add(new int[] { st, ed });
    return ans.ToArray();
  }
}

Solution 2

class Solution:
  def merge(self, intervals: List[List[int]]) -> List[List[int]]:
    intervals.sort()
    ans = [intervals[0]]
    for s, e in intervals[1:]:
      if ans[-1][1] < s:
        ans.append([s, e])
      else:
        ans[-1][1] = max(ans[-1][1], e)
    return ans

class Solution {
  public int[][] merge(int[][] intervals) {
    Arrays.sort(intervals, (a, b) -> a[0] - b[0]);
    List<int[]> ans = new ArrayList<>();
    ans.add(intervals[0]);
    for (int i = 1; i < intervals.length; ++i) {
      int s = intervals[i][0], e = intervals[i][1];
      if (ans.get(ans.size() - 1)[1] < s) {
        ans.add(intervals[i]);
      } else {
        ans.get(ans.size() - 1)[1] = Math.max(ans.get(ans.size() - 1)[1], e);
      }
    }
    return ans.toArray(new int[ans.size()][]);
  }
}

class Solution {
public:
  vector<vector<int>> merge(vector<vector<int>>& intervals) {
    sort(intervals.begin(), intervals.end());
    vector<vector<int>> ans;
    ans.emplace_back(intervals[0]);
    for (int i = 1; i < intervals.size(); ++i) {
      if (ans.back()[1] < intervals[i][0]) {
        ans.emplace_back(intervals[i]);
      } else {
        ans.back()[1] = max(ans.back()[1], intervals[i][1]);
      }
    }
    return ans;
  }
};

func merge(intervals [][]int) (ans [][]int) {
  sort.Slice(intervals, func(i, j int) bool { return intervals[i][0] < intervals[j][0] })
  ans = append(ans, intervals[0])
  for _, e := range intervals[1:] {
    if ans[len(ans)-1][1] < e[0] {
      ans = append(ans, e)
    } else {
      ans[len(ans)-1][1] = max(ans[len(ans)-1][1], e[1])
    }
  }
  return
}

function merge(intervals: number[][]): number[][] {
  intervals.sort((a, b) => a[0] - b[0]);
  const ans: number[][] = [intervals[0]];
  for (let i = 1; i < intervals.length; ++i) {
    if (ans.at(-1)[1] < intervals[i][0]) {
      ans.push(intervals[i]);
    } else {
      ans.at(-1)[1] = Math.max(ans.at(-1)[1], intervals[i][1]);
    }
  }
  return ans;
}

public class Solution {
  public int[][] Merge(int[][] intervals) {
    intervals = intervals.OrderBy(a => a[0]).ToArray();
    var ans = new List<int[]>();
    ans.Add(intervals[0]);
    for (int i = 1; i < intervals.Length; ++i) {
      if (ans[ans.Count - 1][1] < intervals[i][0]) {
        ans.Add(intervals[i]);
      } else {
        ans[ans.Count - 1][1] = Math.Max(ans[ans.Count - 1][1], intervals[i][1]);
      }
    }
    return ans.ToArray();
  }
}

Solution 3

function merge(intervals: number[][]): number[][] {
  intervals.sort((a, b) => a[0] - b[0]);
  const n = intervals.length;
  const res = [];
  let i = 0;
  while (i < n) {
    let [l, r] = intervals[i];
    i++;
    while (i < n && r >= intervals[i][0]) {
      r = Math.max(r, intervals[i][1]);
      i++;
    }
    res.push([l, r]);
  }
  return res;
}