[LeetCode] 435. Non-overlapping Intervals

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原文链接：https://segmentfault.com/a/1190000017060948

本文介绍了一种解决区间重叠问题的算法，旨在通过移除最少数量的区间，使所有剩余区间互不重叠。文章提供了三个示例，展示了算法在不同情况下的应用，并详细解释了如何通过排序和遍历来确定最小移除数。

Problem

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.

Solution

class Solution {
    public int eraseOverlapIntervals(Interval[] intervals) {
        if (intervals == null || intervals.length == 0) return 0;
        //find non-overlapping count, so sort by end
        Arrays.sort(intervals, (a, b)->(a.end-b.end));
        int end = intervals[0].end;
        int count = 1;
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i].start >= end) {
                end = intervals[i].end;
                count++;
            }
        }
        return intervals.length-count;
    }
}