标签：leetcode   算法   

问题描述：

Given a collection of intervals, merge all overlapping intervals.

For example,
Given [1,3],[2,6],[8,10],[15,18],
return [1,6],[8,10],[15,18].

基本思路：

对左边界进行排序，然后对有边界的情况进行分类处理。

代码：

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
 #include <algorithm>
  bool myfunction (Interval A,Interval B)        //C++
     { 
         return (A.start < B.start);
     }
class Solution {
public:
    
    vector<Interval> merge(vector<Interval> &intervals) {
        vector<Interval> result;
        if(intervals.size() == 0)
            return result;
        sort(intervals.begin(),intervals.end(),myfunction);
        
       
        bool haveStart = false;
        int start ;
        int end ;
        for(int i = 0; i < intervals.size(); i++)
        {
            if(haveStart == false && i < intervals.size())
            {
                start = intervals[i].start;
                end   = intervals[i].end;
                haveStart = true;
                continue;
            }
            if(intervals[i].start <= end )
            {
                if(intervals[i].end > end)
                    end = intervals[i].end;
                
            }
            else 
            {
                Interval tmp(start,end);
                result.push_back(tmp);
                start = intervals[i].start;
                end = intervals[i].end;
                
            }
        }
        if(haveStart)
        {
            Interval tmp(start,end);
            result.push_back(tmp);
        }
        return result;
    }
};

[leetcode]Merge Intervals

标签：leetcode   算法   

原文地址：http://blog.csdn.net/chenlei0630/article/details/41488021