0435. Non-overlapping Intervals (M)

Non-overlapping Intervals (M)

题目

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.

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题意

给定一组区间，要求删去最少的区间，使剩余的区间互不重叠。

思路

贪心，类似于规划区间使不重叠的区间最多。先将区间按照左端点升序排序，再依次遍历，如果有两两重叠，则删去右端点较大的那个区间。

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代码实现

Java

class Solution {
    public int eraseOverlapIntervals(int[][] intervals) {
        Arrays.sort(intervals, (int[] a, int[] b) -> a[0] - b[0]);
        int count = 0;
        int i = 0;
        for (int j = 1; j < intervals.length; j++) {
            if (intervals[i][1] > intervals[j][0]) {
                count++;
                i = intervals[i][1] >= intervals[j][1] ? j : i;
            } else {
                i = j;
            }
        }
        return count;
    }
}