【LeetCode & 剑指offer刷题】查找与排序题13:Merge Intervals

【LeetCode & 剑指offer 刷题笔记】目录(持续更新中...)

Merge Intervals

Given a collection of intervals, merge all overlapping intervals.
Example 1:
Input: [[1,3],[2,6],[8,10],[15,18]]
Output: [[1,6],[8,10],[15,18]]
Explanation: Since intervals [1,3] and [2,6] overlaps, merge them into [1,6].
Example 2:
Input: [[1,4],[4,5]]
Output: [[1,5]]
Explanation: Intervals [1,4] and [4,5] are considerred overlapping.

C++
 
/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
//问题:区间合并
/*
方法:将各区间按照第一个元素排序,如果没有重叠,逐个将区间push到结果容器中,
如果有重叠,进行区间合并,修改前一个区间第二个元素即可
*/
class Solution
{
public:
    //自定义函数对象排序(或声明为static类型的函数也可以)
    struct
    {
        bool operator()(Interval& a, Interval& b)//这里用Interval&引用变量比较好,节省时间
        {
            return a.start<b.start;
        }
    } customLess;
       
    vector<Interval> merge(vector<Interval>& ins)
    {
        vector<Interval> res;
        if(ins.empty()) return res;
        sort(ins.begin(), ins.end(),customLess);//按第一个变量排序
        
        res.push_back(ins[0]);
        for(int i=1; i<ins.size(); i++)
        {
            if(ins[i].start > res.back().end) res.push_back(ins[i]); //如果没有重叠时
            else res.back().end = max(res.back().end, ins[i].end); //如果有重叠时,end选较大的那个
        }
       
        return res;
    }
   
};
 

 

posted @ 2019-01-05 20:15  wikiwen  阅读(203)  评论(0编辑  收藏