No.56 Merge Intervals

No.56 Merge Intervals

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].

法一：直接复用之前写的insert函数，依次将区间段插入到结果集中

#include "stdafx.h"
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

struct Interval
{
    int start;
    int end;
    Interval():start(0),end(0) {}
    Interval(int s, int e):start(s),end(e) {}
};
class Solution
{
public:
    vector<Interval> merge(vector<Interval> &intervals)
    {//未排序的区间段数组，进行合并
     //法一：复用之前的insert函数，每次从intervals中取一个区间插入到结果集中
        vector<Interval> res;
        for(int i=0; i<intervals.size(); i++)
            res = insert(res,intervals[i]);
        return res;

    }
    vector<Interval> insert(vector<Interval> &intervals, Interval newInterval)
    {//参考：书
     //从前向后比较，看是否插入。前提：区间段已排序,无重合
     //改进之处：方法没变，但由于insert和erase函数代价有点高，会移动修改，故，不做原地的，空间换时间，直接新建一个好了。
        vector<Interval> res;
        int count = intervals.size();
        if(count == 0)
        {
            res.push_back(newInterval);//防止待插入区间在最后
            return res;
        }

        int index = 0;
        while(index<count)
        {
            if(newInterval.end < intervals[index].start)
            {//当前区间在待插入区间之前，直接插入待插入区间
                res.push_back(newInterval);
                while(index<count)
                {
                    res.push_back(intervals[index]);//剩余元素插入res
                    index++;
                }
                return res;
            }
            else if(newInterval.start > intervals[index].end)
            {//当前区间大于待插入区间，跳过，继续判断
                res.push_back(intervals[index]);
            }
            else
            {//当前区间与待插入区间之间有重合部分
                newInterval.start = min(newInterval.start,intervals[index].start);
                newInterval.end = max(newInterval.end,intervals[index].end);
            }
            index++;
        }
        res.push_back(newInterval);//防止待插入区间在最后
        return res;
    }
};

法二：将数组自定义排序，然后从前向后两两合并

#include "stdafx.h"
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

struct Interval
{
    int start;
    int end;
    Interval():start(0),end(0) {}
    Interval(int s, int e):start(s),end(e) {}
};
class Solution
{
public:
    //自定义的比较函数，为使sort调用，必须是全局或者是静态的，不能是普通成员函数
    static bool compare(Interval v1, Interval v2)
    {
        return v1.start==v2.start ? v1.end<v2.end : v1.start<v2.start;
    }
    vector<Interval> merge(vector<Interval> &intervals)
    {//未排序的区间段数组，进行合并
     //法二：先将区间段数组排序，然后从前向后两两合并
     //参考：http://www.cnblogs.com/ganganloveu/p/4158759.html
        
        vector<Interval> res;//结果集
        int count = intervals.size();
        if(count == 0)
            return res;
        //区间段排序
        sort(intervals.begin(),intervals.end(),compare);
        res.push_back(intervals[0]);

        for(int i=1; i<count; i++)
        {//依次插入当前区间段，将当前区间段与结果集res的最后区间段进行比较，看是否合并——直接修改数据即可
            Interval &back = res.back();//back要是引用，否则不会改变其值!!!
            if(intervals[i].start > back.end)//无重合
                res.push_back(intervals[i]);
            else
            //有重合
                back.end = max(back.end, intervals[i].end);
        }
        return res;
    }
};
int main()
{
    Solution sol;

    Interval data1[] = {Interval(1,3),Interval(2,6),Interval(8,10),Interval(15,18)};
    vector<Interval> test1(data1,data1+4);
//test1
    for(auto &i : test1)
        cout << "["<<i.start << ","<< i.end<<"]";
    cout << endl;
    vector<Interval> res1 = sol.merge(test1);
    for(auto &i : res1)
        cout << "["<<i.start << ","<< i.end<<"]";
    cout << endl;
    cout << endl;

    Interval data2[] = {Interval(3,8),Interval(2,9),Interval(6,7),Interval(8,10),Interval(1,2)};
    vector<Interval> test2(data2,data2+5);
//test2
    for(auto &i : test2)
        cout << "["<<i.start << ","<< i.end<<"]";
    cout << endl;
    vector<Interval> res2 = sol.merge(test2);
    for(auto &i : res2)
        cout << "["<<i.start << ","<< i.end<<"]";
    cout << endl;

    return 0;
}

参考：http://www.cnblogs.com/ganganloveu/p/4158759.html