leetcode 435. Non-overlapping Intervals_会议室问题权值java-优快云博客

本文介绍了一种高效算法，用于解决给定多个区间时，如何通过移除最少数量的区间来避免所有区间重叠的问题。文章详细解释了算法思路，包括如何通过排序和更新右端点来最小化移除操作。

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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.

题目中给出了几个小区间，让去除最少数量的区间，使得剩下的区间没有overlap

overlap: 区间有重叠，哪怕是半个重叠也算 ex:[1,5], [4,7]

思路：
因为区间有左端点，右端点，判断的时候两者兼顾显然比较费时间，比如遇到如下情景
[1,3], [5, 7], [8,9], [2,5]，遇到[2,5]的时候又要从前到后逐个查有没有overlap

可以看到之所以又要从头开始逐个查是因为它的左端点在[5,7], [8,9]的前面，所以只需要让左端点按顺序出现就不需要逐个判断了

然后是右端点
比如刚开始出现了[1, 100]，后面依次是[1,2], [2, 3], [3, 4]
可以看到如果刚开始的区间定义在1～100的范围，那后面所有的小区间都要删除，就不满足题目中的删除最小数量的要求
所以每次都取较小的右端点，才能保证删除数量最小

而左端点因为是按升序排列的，小的左端点会在开始出现，后面不需要再更新左端点

如何判断overlap，只需要右端点 < 下一区间的左端点
题目中说Intervals like [1,2] and [2,3] have borders “touching” but they don’t overlap each other.，所以不需要用<=

    public int eraseOverlapIntervals(int[][] intervals) {
        if (intervals == null || intervals.length <= 1) {
            return 0;
        }
        
        Arrays.sort(intervals, new Comparator() {
           public int compare(Object o1, Object o2) {
               return ((int[])o1)[0] - ((int[])o2)[0];
           } 
        });
        
        int right = intervals[0][1];
        int res = 0;
        
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i][0] < right) {
                res ++;
                right = Math.min(right, intervals[i][1]);
                continue;
            }
            
            if (intervals[i][1] > right) {
                right = intervals[i][1];
            }
            
        }
        
        return res;
    }