[LeetCode] 435. Non-overlapping Intervals

Given an array of intervals intervals where intervals[i] = [starti, endi], return the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

Example 1:

Input: intervals = [[1,2],[2,3],[3,4],[1,3]]
Output: 1
Explanation: [1,3] can be removed and the rest of the intervals are non-overlapping.

Example 2:

Input: intervals = [[1,2],[1,2],[1,2]]
Output: 2
Explanation: You need to remove two [1,2] to make the rest of the intervals non-overlapping.

Example 3:

Input: intervals = [[1,2],[2,3]]
Output: 0
Explanation: You don't need to remove any of the intervals since they're already non-overlapping.

Constraints:

• 1 <= intervals.length <= 105
• intervals[i].length == 2
• -5 * 104 <= starti < endi <= 5 * 104

不重叠的区间。

给定一个区间的集合 intervals ，其中 intervals[i] = [starti, endi] 。返回 需要移除区间的最小数量，使剩余区间互不重叠 。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/non-overlapping-intervals
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思路还是扫描线。既然是找是否有重叠，那么可以根据每个 interval 的结束时间对 input 进行排序。排序之后遍历 intervals，记录不重叠的 interval 一共有几个（记为count），然后用 intervals 的总长度 L - count 得到最后要的结果。跑个例子，

[[1,2],[2,3],[3,4],[1,3]]

按end排序之后的结果是

[ [ 1, 2 ], [ 2, 3 ], [ 1, 3 ], [ 3, 4 ] ]

之后从第二个 interval 开始遍历，如果当前 interval 的 start 大于等于前一个 interval 的 end，说明两者没有 overlap，则 count++，说明这两个 interval 都需要保留。按照这个逻辑，算出最后互不重叠的 intervals 有多少，再用 总数 - count 就得到需要去除的 interval 的数量了。

时间O(nlogn)

空间O(1)

JavaScript实现

/**
 * @param {number[][]} intervals
 * @return {number}
 */
var eraseOverlapIntervals = function(intervals) {
    // corner case
    if (intervals.length === 0) {
        return 0;
    }

    // normal case
    intervals = intervals.sort((a, b) => a[1] - b[1]);
    let end = intervals[0][1];
    let count = 1;
    for (let i = 1; i < intervals.length; i++) {
        if (intervals[i][0] >= end) {
            end = intervals[i][1];
            count++;
        }
    }
    return intervals.length - count;
};

 

Java实现

class Solution {
    public int eraseOverlapIntervals(int[][] intervals) {
        // corner case
        if (intervals.length == 0) {
            return 0;
        }

        // normal case
        Arrays.sort(intervals, (a, b) -> a[1] - b[1]);
        int end = intervals[0][1];
        int count = 1;
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i][0] >= end) {
                end = intervals[i][1];
                count++;
            }
        }
        return intervals.length - count;
    }
}

 

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