LeetCode Non-overlapping Intervals 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: 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.

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给定一系列区间，问最少删除多少个区间能使得剩余区间没有重叠。区间[i,j]和[u,v]重叠是指u<j或者i<v，如果两个区间只是边界重叠比如j==u，则不算重叠。 我们可以使用贪心的策略求最多能保存多少个区间，然后用总数一减就得到最少删除的区间数。 首先对区间先后按start和end从小到大排序，然后从头开始遍历。假设我们上一次保留的区间的end是last_end，如果当前区间的intervals[i].start>=last_end，则说明区间i可以保留，更新last_end=intervals[i].end。如果intervals[i].start<last_end，则当前区间会和上一个区间重叠，需要删除一个区间，为了使留下来的区间更多，肯定要删除end大的区间，令last_end=min(last_end,intervals[i].end)，通过保留end小的区间来间接模拟删除end大的区间。 当得到保留下来的不重叠区间个数ans后，用n-ans就是需要删除的区间数。完整代码如下： [cpp] bool comp(const Interval &i1, const Interval &i2){ return i1.start < i2.start || (i1.start == i2.start && i1.end < i2.end); } class Solution { public: int eraseOverlapIntervals(vector<Interval>& intervals) { if(intervals.empty()) return 0; sort(intervals.begin(), intervals.end(), comp); int ans = 1, n = intervals.size(), last_end = intervals[0].end; for(int i = 1; i < n; ++i){ if(intervals[i].start >= last_end){ ++ans; last_end = intervals[i].end; } else { last_end = min(last_end, intervals[i].end); } } return n – ans; } }; [/cpp] 本代码提交AC，用时9MS。]]>