Leetcode435

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:
You may assume the interval’s end point is always bigger than its start point.
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.

Solution01: 动态规划解法

#include 
#include 

using namespace std;

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

bool compare(const Interval &a, const Interval &b){

    if ( a.start != b.start )
        return a.start < b.start;
    return a.end < b.end;
}

class Solution {
public:
    int eraseOverlapIntervals(vector& intervals) {

        if( intervals.size() == 0)
            return 0;

        sort( intervals.begin(), intervals.end(), compare);

        // memo[i]表示使用intervals[0...i]的区间能构成的最长不重复区间序列长度
        vector memo( intervals.size(), 1 );
        for (int i = 1; i < intervals.size(); i++) {
            //memo[i]
            for (int j = 0; j < i; j++) {
                if ( intervals[i].start >= intervals[j].end)
                    memo[i] = max( memo[i], 1+memo[j]);
            }
        }
        int res = 0;
        for (int k = 0; k < memo.size(); k++) {
            res = max(res, memo[k]);
        }

        return intervals.size() - res;

    }
};

Solution02: 贪心法

#include 
#include 

using namespace std;

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

bool compare(const Interval &a, const Interval &b){

    if ( a.end != b.end )
        return a.end < b.end;
    return a.start < b.start;
}

class Solution {
public:
    int eraseOverlapIntervals(vector& intervals) {

        if( intervals.size() == 0)
            return 0;

        sort( intervals.begin(), intervals.end(), compare);

       int res = 1;
       int pre = 0;
       for (int i = 1; i < intervals.size(); i++) {
            if ( intervals[i].start >= intervals[pre].end ){
                res++;
                pre = i;
            }
        }

        return intervals.size() - res;

    }
};

总结： 动态规划思路：

贪心法思路：