算法分析与设计week07--435. Non-overlapping Intervals

435. Non-overlapping Intervals

Description

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

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.

Analyse

思路：跟场地安排问题类似（要求安排尽量多的活动），贪心算法，以每个元素的end为关键元素排序，end相等的情况下，按start升序排序。贪心规则：尽量安排结束时间早的活动。如果后面的活动与已经安排好的兼容，则加入集合。

查找重叠区间：判断方法是看如果前一个区间的end大于后一个区间的start，那么一定是重复区间，此时结果num加1，删掉一个区间。为了保证我们总体去掉的区间数最小，删掉那个end值较大的区间（使得后面有更宽裕的空间来排入更多的区间）。而在代码中，不需真正的删掉某一个区间，可以用cur指向当前留下来的区间，再与后面的区间作比较。

/** * 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) {        if (intervals.size() == 0) {            return 0;        }        sort(intervals.begin(), intervals.end(), [](Interval a, Interval b){            return a.end < b.end || (a.end == b.end && a.start > b.start);         });        int num = 0;        Interval cur = intervals[0];        for (int i = 1; i < intervals.size(); i++) {            if (intervals[i].start < cur.end) {                num++;                cur = (intervals[i].end > cur.end) ? cur : intervals[i];            } else {                cur = intervals[i];            }        }        return num;    }};

复杂度分析

时间复杂度：O(nlogn)
空间复杂度：O(1)