LeetCode Non-overlapping Intervals

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.

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:

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;
    }
};

本代码提交AC,用时9MS。

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