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[0]);
    for (int i = 1; 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