-
-
Notifications
You must be signed in to change notification settings - Fork 5.7k
/
435. Non-overlapping Intervals.go
76 lines (69 loc) · 1.32 KB
/
435. Non-overlapping Intervals.go
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
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
}
func max(a int, b int) int {
if a > b {
return a
}
return b
}
// 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
}