[算法分析与设计] leetcode 每周一题: Non-overlapping Intervals

题目链接：https://leetcode.com/problems/non-overlapping-intervals/discuss/

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.

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.

思路：

这道题用贪心策略，一开始对区间按照start从小到大排序，然后相邻比较，若相邻覆盖，则必然要舍弃一个，因此舍弃数量+1，至于舍弃哪一个，则看冲突的两个哪个end最大，end越大，越有可能导致被后面区间覆盖，因此end大的要舍弃，如果没有冲突，则继续比较

代码：

class Solution {
public:
int eraseOverlapIntervals(vector<Interval>& intervals) {
auto cmp = [](const Interval & a, const Interval & b) {return a.start < b.start;};
sort(intervals.begin(),intervals.end(),cmp);
int num = 0;
int pre = 0;
for(int i = 1; i < intervals.size(); i++) {
if(intervals[i].start < intervals[pre].end) {
num ++;
if(intervals[i].end < intervals[pre].end)
pre = i;
} else {
pre = i;
}

}
return num;
}
};