[Leetcode] 435. Non-overlapping Intervals 解题报告

题目：

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

Example 2:

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

Example 3:

Input: [ [1,2], [2,3] ]Output: 0Explanation: You don't need to remove any of the intervals since they're already non-overlapping.

思路：

仔细分析可以知道这道题目其实是比较简单的，可以用贪心算法：首先对intervals进行排序，然后遍历：每当遇到重合区域时，我们需要erase一个interval，更新end为两个interval的更靠左者（这是本题思路的关键：如果当前interval的end更靠右，则意味着将要删除当前interval，否则意味着要删除的是前面的某个interval，由于intevals是有序的，所以可推出前面某个interval完全包含了当前interval，因此删除前面的这个interval是安全的）；而如果没有重合，则直接更新end为新的interval的end即可。算法的时间复杂度是O(nlogn)，这是由于排序的缘故所致；空间复杂度则为O(1)。

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:    int eraseOverlapIntervals(vector<Interval>& intervals) {        if (intervals.size() == 0) {            return 0;        }        sort(intervals.begin(), intervals.end(), IntervalComp);        int end = intervals[0].end, ret = 0;        for (int i = 1; i < intervals.size(); ++i) {            if (intervals[i].start < end) {        // need to erase, and we erase the one whose end is larger                ++ret;                end = min(intervals[i].end, end);            }            else {                                 // do not need to erase, so we update the end                end = intervals[i].end;            }        }        return ret;    }private:    struct IntervalCompare {        bool operator() (const Interval &a, const Interval &b) const {            if (a.start < b.start) {                return true;            }            else if (a.start > b.start) {                return false;            }            else {                return a.end < b.end;            }        }    } IntervalComp;};