Leetcode 435. Non-overlapping Intervals

Leetcode 435. Non-overlapping Intervals

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题目描述

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.

输入：一组序列，为活动的起始时间与结束时间
输出：需要剔除的最小活动数量
如：

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.

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.

思路

贪心算法：相当于使不冲突的活动数量最大，考虑下面的贪心-活动-选择算法，该算法需要先对活动以完成时间为key排序。应用于本提时，只需在满足不冲突条件时，使最大活动数量+1，最终通过活动数量-最大活动数量 = 剔除数量即可。其中，排序为O(nlogn)，比较为O(n)，最终复杂度为O(nlogn)。

Greedy-acitivity-selector(s,f)
n <- length[s]
A <- {a1}
i <- 1
for m <- 2 to n
    do if sm >= fi then A <- A U {am}, i <- m
return A

代码

/**
 * 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:
    static bool comp(Interval a, Interval b){
        return a.end<b.end;
    }
    //vector<Interval> sat; //for hold the satisfied intervals
    int ind;//the index of last satisfied activity
    int sats;//the size of satisfied activity

    int eraseOverlapIntervals(vector<Interval>& intervals) {
        int size = intervals.size();

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

        for(int m=0;m<intervals.size();m++){
            if(sats==0/*sat.empty()*/){
                //init
                ind = 0;
                sats = 1;
                //sat.push_back(intervals[m]);
            }else{
                //greed step
                if(intervals[ind].end<=intervals[m].start){
                    ind = m;
                    sats++;
                    //sat.push_back(intervals[m]);
                }
            }
        }
        return size-sats;
    }
};