156. Merge Intervals【easy】

Given a collection of intervals, merge all overlapping intervals.

 

Example

Given intervals => merged intervals:

[                     [
  [1, 3],               [1, 6],
  [2, 6],      =>       [8, 10],
  [8, 10],              [15, 18]
  [15, 18]            ]
]

Challenge

O(n log n) time and O(1) extra space.

解法一：

 /**
  * Definition of Interval:
  * classs Interval {
  *     int start, end;
  *     Interval(int start, int end) {
  *         this->start = start;
  *         this->end = end;
  *     }
  */

 class Solution {
 public:
   /**
    * @param intervals: interval list.
    * @return: A new interval list.
    */
   static bool cmp(const Interval &a, const Interval &b) {
     return (a.start < b.start);
   }

   vector<Interval> merge(vector<Interval>& intervals) {
     vector<Interval> ans;
     if (intervals.empty()) {
       return ans;
     } 

     sort(intervals.begin(), intervals.end(), cmp);
     ans.push_back(intervals[]);
     for (int i = ; i < intervals.size(); i++) {
       if (ans.back().end >= intervals[i].start) {
         ans.back().end = max(ans.back().end, intervals[i].end);
       } else {
         ans.push_back(intervals[i]);
       }
     }
     return ans;
   }
 };

先排序，然后逐个判断是否需要合并