leetcodev 435. Non-overlapping Intervals

Given an array of intervals intervals where intervals[i] = [starti, endi], return the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

Example 1:

Input: intervals = [[1,2],[2,3],[3,4],[1,3]]
Output: 1
Explanation: [1,3] can be removed and the rest of the intervals are non-overlapping.
Example 2:

Input: intervals = [[1,2],[1,2],[1,2]]
Output: 2
Explanation: You need to remove two [1,2] to make the rest of the intervals non-overlapping.
Example 3:

Input: intervals = [[1,2],[2,3]]
Output: 0
Explanation: You don’t need to remove any of the intervals since they’re already non-overlapping.

Constraints:

1 <= intervals.length <= 105
intervals[i].length == 2
-5 * 104 <= starti < endi <= 5 * 104

Greedy Algorithms

bool compare(vector<int> & a, vector<int> & b);
class Solution {
public:
    int eraseOverlapIntervals(vector<vector<int>>& intervals) {
        sort(intervals.begin(), intervals.end(),compare );
        int maxNum=0;
        int end =intervals[0][0];

        for(int i=0;i<intervals.size();i++)
        {
            if(end<= intervals[i][0])
            {
                maxNum++;
                end = intervals[i][1];
                //cout<< " [ "<< curinterval[0] << " , "<< curinterval[1]<<" ]"<< endl;
            }
        }
        return  intervals.size() - maxNum;
    }

    
};

bool compare(vector<int> & a, vector<int> & b)
{
    if(a[1]==b[1])
    {
        return (a[1] - a[0])< (b[1] - b[0]);
    }
    return a[1]<b[1];
}

bool compare(vector<int> & a, vector<int> & b);
class Solution {
public:
    int eraseOverlapIntervals(vector<vector<int>>& intervals) {
        sort(intervals.begin(), intervals.end(),compare );
        int maxNum=0;
        int begin =intervals[0][1];

        for(int i=0;i<intervals.size();i++)
        {
            if(begin>= intervals[i][1])
            {
                maxNum++;
                begin = intervals[i][0];
                cout<< " [ "<<  intervals[i][0] << " , "<<  intervals[i][1]<<" ]"<< endl;
            }
        }
        return  intervals.size() - maxNum;
    }

    
};

bool compare(vector<int> & a, vector<int> & b)
{
    return a[0]>b[0];
}

We shall complete the process of developing a greedy solution by converting the recursive algorithmto an iterative one.

参考文献
https://edutechlearners.com/download/Introduction_to_algorithms-3rd%20Edition.pdf

https://leetcode.com/problems/non-overlapping-intervals/solutions/3785409/beat-s-100-c-java-python-beginner-friendly