435. Non-overlapping Intervals

Given a collection of intervals, find the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

Note:

1. You may assume the interval's end point is always bigger than its start point.
2. Intervals like [1,2] and [2,3] have borders "touching" but they don't overlap each other.

 

Example 1:

Input: [ [1,2], [2,3], [3,4], [1,3] ]

Output: 1

Explanation: [1,3] can be removed and the rest of intervals are non-overlapping.

 

Example 2:

Input: [ [1,2], [1,2], [1,2] ]

Output: 2

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

 

Example 3:

Input: [ [1,2], [2,3] ]

Output: 0

Explanation: You don't need to remove any of the intervals since they're already non-overlapping.

 

Approach #1: C++. [stack][8ms]

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
class Solution {
public:
    int eraseOverlapIntervals(vector<Interval>& intervals) {
        int size = intervals.size();
        if (size == 0) return 0;
        sort(intervals.begin(), intervals.end(), [](Interval& a, Interval& b) {
        if (a.end == b.end) return a.start < b.start;
        return a.end < b.end;});
        
        stack<Interval> temp;
        temp.push(intervals[0]);
        int ans = 0;
        for (int i = 1; i < size; ++i) {
            if (temp.top().end <= intervals[i].start)
                temp.push(intervals[i]);
            else ans++;
        }
        return ans;
    }
};

　　

　　

Approach #2: C++. [4ms]

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
class Solution {
public:
    int eraseOverlapIntervals(vector<Interval>& intervals) {
        int len = intervals.size();
        if (len==0) return 0;        
        sort(intervals.begin(), intervals.end(), [](const Interval &a, const Interval &b) {
            return a.start < b.start;
        });
        int end = intervals[0].end;
        int ans = 0;
        for(int i=1; i<len; i++) {
            auto v = intervals[i];
            if (v.start< end) { // overlap
                ans++;
                // shorter end survives
                end = min(end, v.end);
            } else {
                // update new end
                end = v.end;
            }
        }
            
        
        return ans;
    }
};