文章目录
- 一、题目
- 二、题解
一、题目
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
二、题解
class Solution {
public:static bool cmp(vector<int>& a,vector<int>& b){return a[0] < b[0];}vector<vector<int>> merge(vector<vector<int>>& intervals) {int n = intervals.size();sort(intervals.begin(),intervals.end(),cmp);vector<vector<int>> res;res.push_back(intervals[0]);for(int i = 1;i < n;i++){//更新右边界值if(res.back()[1] >= intervals[i][0]){res.back()[1] = max(res.back()[1],intervals[i][1]);}else res.push_back(intervals[i]);}return res;}
};