【LeetCode】435. Non-overlapping Intervals (Medium) (JAVA)
Subject address: https://leetcode.com/problems/non-overlapping-intervals/
Title description:
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:
- 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.
General idea
Given a set of intervals, find the minimum number of intervals that need to be removed so that the remaining intervals do not overlap each other.
note:
- It can be considered that the end of an interval is always greater than its starting point.
- The boundaries of the intervals [1,2] and [2,3] "touch" each other, but do not overlap each other.
Problem-solving method
- This question is sorted first. After sorting, delete the overlapping elements behind and before, and then count the number of deleted elements
- The focus is on how to sort, such as: [[1,2],[2,3],[3,4],[1,3]], how do you choose to delete elements?
- If you want to delete as few elements as possible, you need to delete the elements with a large span. How to find the elements with a large span? 1. The beginning is the same, but the end is larger, the span is longer; 2. The end is the same, and the beginning is smaller, the span is longer
- The first way of sorting is easy to cite the opposite. For example: [[1,9],[2,3],[3,4],[4,5]], where [1,9] can only delete all in front The latter element, definitely not
- Use the second method: the end is the same, and the beginning is smaller, the span is longer. Sort according to the end, the end is the same, and the beginning is the smallest [[1,9],[2,3],[3,4],[4,5]] --> [[2,3], [3,4],[4,5],[1,9]]
class Solution {
public int eraseOverlapIntervals(int[][] intervals) {
Arrays.sort(intervals, (a, b) -> (a[1] - b[1] == 0 ? -a[0] + b[0] : a[1] - b[1]));
int res = 0;
int index = 0;
for (int i = 1; i < intervals.length; i++) {
if (intervals[i][0] < intervals[index][1]) {
res++;
} else {
index = i;
}
}
return res;
}
}
Execution time: 4 ms, defeating 51.09% of Java users
Memory consumption: 38.4 MB, defeating 59.37% of Java users
