1 /**
2 * 普通递归，效率低下，大部分值都被重复计算多遍
3 * 时间复杂度：O(2^n)
4 *
5 * n=50时，值：12586269025，花费36s
6 *
7 */
8 @Test
9 public void recursive() {
10 long st = System.currentTimeMillis();
11 System.out.println(testRecursive(50));
12 System.out.println("cost Seconds: " + (System.currentTimeMillis() - st) / 1000);
13 }
14
15 private long testRecursive(int n) {
16 if (n == 1 || n == 2) {
17 return 1;
18 }
19 return testRecursive(n - 1) + testRecursive(n - 2);
20 }
21
22 /**
23 * 递归 + cache
24 * 时间复杂度：O(n)
25 * 空间复杂度：O(n)
26 *
27 * n=50时，值：12586269025，花费0s
28 */
29
30 @Test
31 public void cache() {
32 long st = System.currentTimeMillis();
33 Map<Long, Long> cache = new HashMap<>();
34 System.out.println(testCache(50L, cache));
35 System.out.println("cost Seconds: " + (System.currentTimeMillis() - st) / 1000);
36 }
37
38 private Long testCache(Long n, Map<Long, Long> cache) {
39 if (n == 1 || n == 2) {
40 return 1L;
41 }
42 if (cache.containsKey(n)) {
43 return cache.get(n);
44 }
45
46 long val = testCache(n - 1, cache) + testCache(n - 2, cache);
47 cache.put(n, val);
48 return val;
49 }
50
51 /**
52 * 自底向上 + cache
53 * 时间复杂度：O(n)
54 * 空间复杂度：O(n)
55 *
56 * n=50时，值：12586269025，花费0s
57 */
58 @Test
59 public void dp() {
60 long st = System.currentTimeMillis();
61 System.out.println(testDp(50));
62 System.out.println("cost Seconds: " + (System.currentTimeMillis() - st) / 1000);
63 }
64
65 private long testDp(int n) {
66 Map<Integer, Long> dpCache = new HashMap<>();
67 if (n == 1 || n == 2) {
68 return 1;
69 }
70 dpCache.put(1, 1L);
71 dpCache.put(2, 1L);
72 for (int i = 3; i <= n; i++) {
73 dpCache.put(i, dpCache.get(i - 1) + dpCache.get(i - 2));
74 }
75 return dpCache.get(n);
76 }
77
78 /**
79 * 自底向上 + 空间复杂度为O(1)
80 * 时间复杂度：O(n)
81 * 空间复杂度：O(1)
82 *
83 *
84 * n=50时，值：12586269025，花费0s
85 */
86 @Test
87 public void dp2() {
88 long st = System.currentTimeMillis();
89 System.out.println(testDp2(50));
90 System.out.println("cost Seconds: " + (System.currentTimeMillis() - st) / 1000);
91 }
92
93 private long testDp2(int n) {
94 if (n == 1 || n == 2) {
95 return 1;
96 }
97 long pre = 1;
98 long after = 1;
99 long tmp;
100 for (int i = 3; i <= n; i++) {
101 tmp = pre + after;
102 pre = after;
103 after = tmp;
104 }
105 return after;
106 }
