Leetcode 435. Non-overlapping Intervals

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:
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.

Example1:

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.

Example2:

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.

Example3:

Input: [ [1,2], [2,3] ]

Output: 0

Explanation: You don't need to remove any of the intervals since they're already non-overlapping.

思路

先对区间进行排序，然后遍历所有区间，尽量去掉长区间，则保留的区间数目一定最多。

代码

bool cmp(Interval a, Interval b) {
    return a.start < b.start; 
}

class Solution {
public:
    int eraseOverlapIntervals(vector& intervals) {
        sort(intervals.begin(), intervals.end(), cmp);
        int res = 0, pre = 0; //res记录保留区间的个数，pre记录每次需要判断是否重叠的起始区间
        for (int i = 1; i < intervals.size(); i++) {
            if (intervals[i].start < intervals[pre].end) {
                res++; //当前区间起始值小于pre区间的末尾保留上一个区间
                if (intervals[i].end < intervals[pre].end) pre = i;
                //这个的意思是从集合中去掉的元素是pre（因为pre太长）
            }
            else pre = i; //区间i和区间pre已经不重叠，需要判断重叠的起始区间pre后移
        }
        return res;
    }
};