0435. Non Overlapping Intervals

435. Non-overlapping Intervals #

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.

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.

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: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.

题目大意 #

给定一个区间的集合,找到需要移除区间的最小数量,使剩余区间互不重叠。

注意:

  1. 可以认为区间的终点总是大于它的起点。
  2. 区间 [1,2] 和 [2,3] 的边界相互“接触”,但没有相互重叠。

解题思路 #

  • 给定一组区间,问最少删除多少个区间,可以让这些区间之间互相不重叠。注意,给定区间的起始点永远小于终止点。[1,2] 和 [2,3] 不叫重叠。
  • 这一题可以反过来考虑,给定一组区间,问最多保留多少区间,可以让这些区间之间相互不重叠。先排序,判断区间是否重叠。
  • 这一题一种做法是利用动态规划,模仿最长上升子序列的思想,来解题。
  • 这道题另外一种做法是按照区间的结尾进行排序,每次选择结尾最早的,且和前一个区间不重叠的区间。选取结尾最早的,就可以给后面留出更大的空间,供后面的区间选择。这样可以保留更多的区间。这种做法是贪心算法的思想。

代码 #


package leetcode

import (
	"sort"
)

// 解法一 DP O(n^2) 思路是仿造最长上升子序列的思路
func eraseOverlapIntervals(intervals [][]int) int {
	if len(intervals) == 0 {
		return 0
	}
	sort.Sort(Intervals(intervals))
	dp, res := make([]int, len(intervals)), 0
	for i := range dp {
		dp[i] = 1
	}
	for i := 1; i < len(intervals); i++ {
		for j := 0; j < i; j++ {
			if intervals[i][0] >= intervals[j][1] {
				dp[i] = max(dp[i], 1+dp[j])
			}
		}
	}
	for _, v := range dp {
		res = max(res, v)
	}
	return len(intervals) - res
}

// Intervals define
type Intervals [][]int

func (a Intervals) Len() int {
	return len(a)
}
func (a Intervals) Swap(i, j int) {
	a[i], a[j] = a[j], a[i]
}
func (a Intervals) Less(i, j int) bool {
	for k := 0; k < len(a[i]); k++ {
		if a[i][k] < a[j][k] {
			return true
		} else if a[i][k] == a[j][k] {
			continue
		} else {
			return false
		}
	}
	return true
}

// 解法二 贪心 O(n)
func eraseOverlapIntervals1(intervals [][]int) int {
	if len(intervals) == 0 {
		return 0
	}
	sort.Sort(Intervals(intervals))
	pre, res := 0, 1

	for i := 1; i < len(intervals); i++ {
		if intervals[i][0] >= intervals[pre][1] {
			res++
			pre = i
		} else if intervals[i][1] < intervals[pre][1] {
			pre = i
		}
	}
	return len(intervals) - res
}


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