力扣算法题—056合并区间

给出一个区间的集合，请合并所有重叠的区间。

示例 1:

输入: [[1,3],[2,6],[8,10],[15,18]]
输出: [[1,6],[8,10],[15,18]]
解释: 区间 [1,3] 和 [2,6] 重叠, 将它们合并为 [1,6].

示例 2:

输入: [[1,4],[4,5]]
输出: [[1,5]]
解释: 区间 [1,4] 和 [4,5] 可被视为重叠区间。

#include "000库函数.h"


 struct Interval {
     int start;
     int end;
     Interval() : start(0), end(0) {}
     Interval(int s, int e) : start(s), end(e) {}
 };

 //第一印象就是排序再组合合并
//失败，因为给的范围不是按顺序的
//class Solution {
//public:
//    vector<Interval> merge(vector<Interval>& intervals) {
//        if (intervals.empty())return {};
//        vector<Interval>res;
//        Interval p;
//        p.start = intervals[0].start;
//        for (int i = 1; i < intervals.size(); ++i) {
//            if ((intervals[i].start - intervals[i-1].end) < 1)
//                continue;
//            p.end = intervals[i-1].end;
//            res.push_back(p);
//            p.start = intervals[i].start;
//        }
//        p.end = intervals[intervals.size()-1].end;
//        res.push_back(p);
//        return res;
//    }
//};

//仍然是先排序在合并的思想
//但是改进了，将头尾分开排序
//20ms
 class Solution {
 public:
     vector<Interval> merge(vector<Interval>& intervals) {
         if (intervals.empty())return {};
         vector<Interval>res;
         vector<int>s, e;
         for (int i = 0; i < intervals.size(); ++i) {
             s.push_back(intervals[i].start);
             e.push_back(intervals[i].end);
         }
         sort(s.begin(), s.end());//头排序
         sort(e.begin(), e.end());//尾排序
         Interval p;
         p.start = s[0];
         for (int i = 1; i < s.size(); ++i) {
             if ((s[i] - e[i - 1]) < 1)continue;//比较大小
             p.end = e[i - 1];
             res.push_back(p);
             p.start = s[i];
         }
         p.end = e[s.size() - 1];
         res.push_back(p);
         return res;
     }
 };

 //将上述方法改进，使用自定义排序函数
 class Solution {
 public:
     vector<Interval> merge(vector<Interval>& intervals) {
         if (intervals.empty()) return {};
         sort(intervals.begin(), intervals.end(), [](Interval &a, Interval &b) {return a.start < b.start; });
         vector<Interval> res{ intervals[0] };
         for (int i = 1; i < intervals.size(); ++i) {
             if (res.back().end < intervals[i].start) {
                 res.push_back(intervals[i]);
             }
             else {
                 res.back().end = max(res.back().end, intervals[i].end);
             }
         }
         return res;
     }
 };


//这道题还有另一种解法，由于插入的过程中也有合并的操作，所以我们可以建立一个空的集合，
//然后把区间集的每一个区间当做一个新的区间插入结果中，也可以得到合并后的结果，

 class Solution {
 public:
     vector<Interval> merge(vector<Interval>& intervals) {
         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) {
         vector<Interval> res;
         int n = intervals.size(), cur = 0;
         for (int i = 0; i < n; ++i) {
             if (intervals[i].end < newInterval.start) {
                 res.push_back(intervals[i]);
                 ++cur;
             }
             else if (intervals[i].start > newInterval.end) {
                 res.push_back(intervals[i]);
             }
             else {
                 newInterval.start = min(newInterval.start, intervals[i].start);
                 newInterval.end = max(newInterval.end, intervals[i].end);
             }
         }
         res.insert(res.begin() + cur, newInterval);
         return res;
     }
 };



void T056() {
    Solution s;
    vector<Interval>v;
    vector<int>nums;
    Interval p;
    nums = { 1,3,2,6,8,10,15,18 };
    for (int i = 0; i < nums.size(); i += 2) {
        p.start = nums[i];
        p.end = nums[i + 1];
        v.push_back(p);
    }
    v = s.merge(v);
    for (auto a : v)
        cout << a.start << ", " << a.end << ";  ";
    cout << endl;

    v.resize(0);
    nums = { 1,4,4,5 };
    for (int i = 0; i < nums.size(); i += 2) {
        p.start = nums[i];
        p.end = nums[i + 1];
        v.push_back(p);
    }
    v = s.merge(v);
    for (auto a : v)
        cout << a.start << ", " << a.end << ";  ";
    cout << endl;

    v.resize(0);
    nums = { 1,4,0,4 };
    for (int i = 0; i < nums.size(); i += 2) {
        p.start = nums[i];
        p.end = nums[i + 1];
        v.push_back(p);
    }
    v = s.merge(v);
    for (auto a : v)
        cout << a.start << ", " << a.end << ";  ";
    cout << endl;

}