思路

对数组排序，按照左端点从小到大排序。初始化Merged，将第一个区间放入。遍历intervals ,如果当前区间的左端点比merged最后一个区间的右端点大，不重合，直接将该区间加入最后，否则，重合，更新最后一个区间的右端点。

Code


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

        // 合并区间 
        
        int n = intervals.size() ; 

        sort(intervals.begin() , intervals.end()) ; 
        vector<vector<int>> merged ; 

        merged.push_back(intervals[0]) ;
        int r = intervals[0][1] ; 
        
        for(int i = 1 ; i<intervals.size() ; i++) {
            int start = intervals[i][0] ; 
            int end = intervals[i][1] ; 
            // 如果当前的左端点小于之前的右，不重合
            if(start > r) {
                merged.push_back({start,end}) ;
            }else{
                // 否则，重合，更新右端点
                merged.back()[1] = max(merged.back()[1] ,end) ;
            }
            r = merged.back()[1] ; 
            
        }
        return merged ; 
    }
};

class Solution:
    def merge(self, intervals: List[List[int]]) -> List[List[int]]:
        # 合并区间
        intervals.sort()  # Sort the intervals based on their start points
        merged = [intervals[0]]
        r = intervals[0][1]

        for i in range(1, len(intervals)):
            start, end = intervals[i]
            # If the current interval's start is greater than the previous interval's end, they don't overlap
            if start > r:
                merged.append([start, end])
            else:
                # Otherwise, they overlap, update the end point of the merged interval
                merged[-1][1] = max(merged[-1][1], end)
            r = merged[-1][1]

        return merged