| 统计删除几个区间使数组变为无重叠的区间 Non | 您所在的位置：网站首页 › Countif区间统计 › 统计删除几个区间使数组变为无重叠的区间 Non |
统计删除几个区间使数组变为无重叠的区间 Non
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问题： 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: 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.解决： ① 先排序，再删除。根据每个区间的start来做升序排序，然后我们开始要查找重叠区间，判断方法是看如果前一个区间的end大于后一个区间的start，那么一定是重复区间，此时我们结果res自增1，我们需要删除一个，那么此时我们究竟该删哪一个呢，为了保证我们总体去掉的区间数最小，我们去掉那个end值较大的区间。 /** * Definition for an interval. * public class Interval { * int start; * int end; * Interval() { start = 0; end = 0; } * Interval(int s, int e) { start = s; end = e; } * } */ class Solution { //5ms public int eraseOverlapIntervals(Interval[] intervals) { Arrays.sort(intervals, new Comparator() { public int compare(Interval o1, Interval o2) {//按照start升序排列 return o1.start - o2.start; } }); int count = 0;//记录要删除的个数 int last = 0; int len = intervals.length; for (int i = 1;i < len;i ++){ if (intervals[i].start < intervals[last].end){ count ++; if (intervals[i].end < intervals[last].end) last = i; }else { last = i; } } return count; } } |
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