Leetcode:435. Non-overlapping Intervals (week 9)

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:
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.

解题思路及算法分析

题意是要移除重叠区间，结果返回为移除的最小区间数。本题可用贪心算法解决，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。本题跟“只有一个场地，要安排尽量多的活动“类似。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。时间复杂度为O(n).
注意的是排序时调用sort要自定义排序条件

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 cmp(Interval a, Interval b) {        if (a.end <= b.end) {            if (a.end < b.end) {                return true;            }            return a.start <= b.start;        }        return false;    }    int eraseOverlapIntervals(vector<Interval>& intervals) {        int size = intervals.size();        if(size == 0) {return 0;}        int count = 0;//移除的区间数        sort(intervals.begin(),intervals.end(),cmp);        int g = intervals[0].end;//必须是排完序后的第一个区间        for (int i = 1; i < size; i++) {            if (intervals[i].start < g) {                count++;            } else g = intervals[i].end;        }        return count;    }};

或者可以将start排序，代码如下

/** * 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;        int count = 0;//移除的区间数        int g = 0;        sort(intervals.begin(),intervals.end(),[](Interval a, Interval b) {            return a.start < b.start;        });        for (int i = 1; i < size; i++) {            if (intervals[i].start < intervals[g].end) {                count++;                if (intervals[i].end < intervals[g].end) g = i;            } else {                g = i;            }        }        return count;    }};