0%

算法-综合

1. LRU 缓存机制(中等)

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
type LRUCache struct {
Capacity int
Map map[int]int
List []int
}


func Constructor(capacity int) LRUCache {
return LRUCache{
Capacity : capacity,
Map : make(map[int]int, 0),
List : make([]int, 0),
}
}

func DelKey(a []int, key int) []int{
for i := 0; i < len(a); i++ {
if a[i] == key {
return append(a[:i], a[i+1:]...)
}
}
return a[1:len(a)]
}

func (this *LRUCache) Get(key int) int {
if val, ok := this.Map[key]; ok{
this.List = DelKey(this.List, key)
this.List = append(this.List, key)
return val
}
return -1
}


func (this *LRUCache) Put(key int, value int) {

if _, ok := this.Map[key]; ok{
this.Map[key] = value
this.List = DelKey(this.List, key)
this.List = append(this.List, key)
return
}

if len(this.Map) >= this.Capacity{
delKey := this.List[0]
this.List = DelKey(this.List, key)
delete(this.Map, delKey)
}

this.Map[key] = value
this.List = append(this.List, key)
}

2. 顺时针打印矩阵

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
func spiralOrder(matrix [][]int) []int {
if len(matrix) == 0 || len(matrix[0]) == 0{
return []int{}
}

res := []int{}
l := 0
r := len(matrix[0]) - 1
t := 0
b := len(matrix) - 1

for {
// 左到右
for i := l ; i <= r; i++ {
res = append(res, matrix[t][i])
}
t++
if t > b {
break
}

// 右到下
for i := t; i <= b; i++ {
res = append(res, matrix[i][r])
}

r--
if l > r {
break
}

// 下到左
for i := r; i >= l; i-- {
res = append(res, matrix[b][i])
}
b--
if t > b {
break
}


// 左到上
for i := b; i >= t; i-- {
res = append(res, matrix[i][l])
}
l++
if l > r {
break
}
}

return res
}

给作者打赏,可以加首页微信,咨询作者相关问题!