[LeetCode] 435. 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. 

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.

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.

不重叠的区间。

题意是给了一组intervals，求至少需要删除几个interval就能使得最后的结果集中没有重叠。这题又是用到扫描线的思想。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就得到需要剪掉的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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