logseq_notes/pages/OJ notes/pages/Leetcode Merge-Intervals.md

Leetcode Merge-Intervals

id:: 642edfd5-4367-495d-b836-510b0deaa3bf

2022-09-01 14:04

Algorithms:

#algorithm #sort

Data structures:

#DS #array

Difficulty:

#coding_problems #difficulty_medium

Additional tags:

#leetcode

Revisions:

N/A

Links:

• Link to problem

• Solution ans Explanation

  ---

Problem

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.

• Examples

• **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

Thoughts

[!summary] This is a generic array problem.

Situations to consider:

• The intervals can be unordered

• The first interval

• [0, 3] [0, 1] are adjacent and overlapped.

  To solve the situations, sort first, and use max function to solve the 3rd solution.

• Solution

  class Solution {
  public:
    vector<vector<int>> merge(vector<vector<int>> &intervals) {
      // sort first, so that data are continious
      vector<vector<int>> ans;
  
      sort(intervals.begin(), intervals.end());
  
      for (auto interval : intervals) {
        if (ans.empty() || ans.back()[1] < interval[0]) {
          ans.push_back(interval);
        } else {
          ans.back()[1] = max(interval[1], ans.back()[1]);
        }
      }
  
      return ans;
    }
  };