435. Non-overlapping Intervals

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

My Solutions:

用greedy的思想，因为要尽可能多的保留intervals， sort the intervals by the end time and keep the ones that end early

class Solution {
    public int eraseOverlapIntervals(int[][] intervals) {

        Arrays.sort(intervals, (a, b) -> Integer.compare(a[1], b[1]));
        int end = intervals[0][1], start = intervals[0][0], count = 0;

        for (int i = 1; i < intervals.length; i++) {
            int[] interval = intervals[i];
            int currStart = interval[0], currEnd = interval[1];
            if (currStart < end) count++; // 这种情况一定要删除一个interval
            else { // 更新两个时间
                start = currStart;
                end = currEnd;
            }
        }
        return count;
    }
}