README_EN.md

435. Non-overlapping Intervals

中文文档

Description

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

Solutions

Greedy.

Python3

class Solution:
    def eraseOverlapIntervals(self, intervals: List[List[int]]) -> int:
        if not intervals:
            return 0
        intervals.sort(key=lambda x: x[1])
        cnt, end = 0, intervals[0][1]
        for interval in intervals[1:]:
            if interval[0] >= end:
                end = interval[1]
            else:
                cnt += 1
        return cnt

Java

class Solution {
    public int eraseOverlapIntervals(int[][] intervals) {
        if (intervals == null || intervals.length == 0) {
            return 0;
        }
        Arrays.sort(intervals, Comparator.comparingInt(a -> a[1]));
        int end = intervals[0][1], cnt = 0;
        for (int i = 1; i < intervals.length; ++i) {
            if (intervals[i][0] >= end) {
                end = intervals[i][1];
            } else {
                ++cnt;
            }
        }
        return cnt;
    }
}

C++

class Solution {
public:
    int eraseOverlapIntervals(vector<vector<int>> &intervals) {
        if (intervals.empty())
        {
            return 0;
        }
        sort(intervals.begin(), intervals.end(), [](const auto &a, const auto &b)
             { return a[1] < b[1]; });
        int ed = intervals[0][1], cnt = 0;
        for (int i = 1; i < intervals.size(); ++i)
        {
            if (ed <= intervals[i][0])
            {
                ed = intervals[i][1];
            }
            else
            {
                ++cnt;
            }
        }
        return cnt;
    }
};

Go

func eraseOverlapIntervals(intervals [][]int) int {
	if intervals == nil || len(intervals) == 0 {
		return 0
	}
	sort.Slice(intervals, func(i, j int) bool {
		return intervals[i][1] < intervals[j][1]
	})
	end, cnt := intervals[0][1], 0
	for i := 1; i < len(intervals); i++ {
		if intervals[i][0] >= end {
			end = intervals[i][1]
		} else {
			cnt++
		}
	}
	return cnt
}

...