Merge Intervals

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

Code

類似 Insert Interval。關鍵都在於看後者的開頭是否小於前者的尾巴，再去做處理。

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

class Solution {
public:
    vector<vector<int>> merge(vector<vector<int>>& intervals) {
        int i = 1, n = intervals.size();
        sort(intervals.begin(), intervals.end());
 
        vector<vector<int>> res;
        res.push_back(intervals[0]);
 
        while(i < n) {
            while(i < n && res.back()[1] < intervals[i][0]) {
                res.push_back(intervals[i]);
                i++;
            }
 
            while(i < n && res.back()[1] >= intervals[i][0]) {
                res.back()[0] = min(res.back()[0], intervals[i][0]);
                res.back()[1] = max(res.back()[1], intervals[i][1]);
                i++;
            }
        }
        return res;
    }
};

Source

• Merge Intervals - LeetCode