# 【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
/*

*/
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