Leetcode56. consolidation interval

One ： The analects of Confucius

It feels hard to do this , Because there are few falls , So many things are simply exquisite , Speed , As one can imagine , The results were all bad .

Two ： subject

3、 ... and ： Upper code

// class Solution {
    
// public:

// static bool cmp(const vector<int>&v1, const vector<int>& v2) {
    
// return v1[1] < v2[1];
// }
// vector<vector<int>> merge(vector<vector<int>>& intervals) {
    
// /**
//  Ideas :1. This question is similar to the non overlapping interval 
// 2. Let's start with the left bound of the interval array in ascending order 
// 3. Then judge the right boundary of the first group of elements and compare it with the left boundary of the second group of elements , If it's larger, merge , Simultaneous updating 
//  Right border .
// */
// vector<vector<int> >ans;
// int flag = 0;
// if(intervals.size() == 1) {
    
// ans.push_back(intervals[0]);
// }

// sort(intervals.begin(),intervals.end(),cmp);
// int end = intervals[0][1];// The right boundary of the first set of elements 
// int start = intervals[0][0];// The left boundary of the first set of elements 

// if(intervals.size() > 1 && intervals[1][0] > intervals[0][1]) {// Special treatment of the first group of elements 
// ans.push_back(intervals[0]);
// }

// for(int i = 1; i < intervals.size(); i++) {
    
            
// while(i < intervals.size() && end >= intervals[i][0]) {// Merge if it is larger than its left boundary 
// end = intervals[i][1];
// start = min(start,intervals[i][0]);
// i++;
// flag = 1;
// if(i == intervals.size()) ans.push_back({start,end});
// }

// if(i < intervals.size() && flag == 1){
    
// ans.push_back({start,end});
// flag = 0;
// }

// if(i < intervals.size() && end < intervals[i][0]) {
    
// ans.push_back(intervals[i]);
// }
            
// }
// return ans;
// }
// };

class Solution {
    
public:

    static bool cmp(const vector<int>&v1, const vector<int>& v2) {
    
        return v1[0] < v2[0];
    }
    vector<vector<int>> merge(vector<vector<int>>& intervals) {
    
        /**  Ideas :1. This question is similar to the non overlapping interval  2. Let's start with the left bound of the interval array in ascending order  3. Then judge the left boundary of the second group of elements and compare it with the right boundary of the first group of elements , If it's larger, merge , Simultaneous updating   Right border . 4.intervals[i-1][1] >= inteavals[i][0]; Then we can judge the overlap , Because we do it according to the left bound of the array   ascend , So  intervals[i][0] > intervals[i-1][0],  Then there must be overlap . */
        vector<vector<int> >ans;
        int flag = 0;
       
        sort(intervals.begin(),intervals.end(),cmp);

        for(int i = 1; i < intervals.size(); i++) {
    
            int start = intervals[i-1][0];// there i-1, Dug a hole for our last set of elements , If the last set of elements is not merged 
            int end = intervals[i-1][1];  // there i-1 Will separate them .

            while(i < intervals.size() && intervals[i][0] <= end) {
    // The right boundary of the previous element is greater than the left boundary of the current element 
                end = max(end,intervals[i][1]);// Because maybe the right boundary is not in ascending order 
                if(i == intervals.size()-1) flag = 1;
                i++;
            }
            ans.push_back({
    start,end});
        }
        if(flag == 0) ans.push_back(intervals[intervals.size()-1]);
        return ans;
    }
};

For two hours , I can't write it myself , Dystocia , You can't give birth even if you're tired to death , I used by the right boundary of the array Ascending processing , I thought it was similar to the problem I had done before , But it's not , We deal with the right boundary if people are in ascending order , So when there is a set of ranges covering all array ranges
We'll have trouble giving birth , But you won't encounter this problem with the left boundary .

What kind of feelings are firm ,
From university to society ,
To get married again ？
Do ordinary things ,
We are ordinary people
But I always feel like a Superhero
Ha ha ha Nonsense over Good night, Come on, stranger .