1 """
2 Given two strings text1 and text2, return the length of their longest common subsequence.
3 A subsequence of a string is a new string generated from the original string with some characters(can be none) deleted without changing the relative order of the remaining characters. (eg, "ace" is a subsequence of "abcde" while "aec" is not). A common subsequence of two strings is a subsequence that is common to both strings.
4 If there is no common subsequence, return 0.
5 Example 1:
6 Input: text1 = "abcde", text2 = "ace"
7 Output: 3
8 Explanation: The longest common subsequence is "ace" and its length is 3.
9 Example 2:
10 Input: text1 = "abc", text2 = "abc"
11 Output: 3
12 Explanation: The longest common subsequence is "abc" and its length is 3.
13 Example 3:
14 Input: text1 = "abc", text2 = "def"
15 Output: 0
16 Explanation: There is no such common subsequence, so the result is 0.
17 """
18
19 """
20 提交显示wrong answer，但在IDE是对的
21 其实这个代码很冗余，应该是 超时
22 """
23 class Solution1:
24 def longestCommonSubsequence(self, text1, text2):
25 res1, res2 = 0, 0
26 i, j = 0, 0
27 while i < len(text1) and j < len(text2):
28 if text1[i] == text2[j]:
29 res1 += 1
30 i += 1
31 j += 1
32 else:
33 if len(text1) - i < len(text2) - j: #第二遍循环是为了处理这个
34 i += 1
35 else:
36 j += 1
37 i, j = 0, 0
38 while i < len(text1) and j < len(text2):
39 if text1[i] == text2[j]:
40 res2 += 1
41 i += 1
42 j += 1
43 else:
44 if len(text1) - i > len(text2) - j:
45 i += 1
46 else:
47 j += 1
48 res = max(res1, res2)
49 return res
50
51
52 """
53 经典的动态规划，用一个二维数组，存当前的结果
54 如果值相等：dp[i][j] = dp[i-1][j-1] + 1
55 如果值不等：dp[i][j] = max(dp[i-1][j], dp[i][j-1]) 左边和上边的最大值
56 在矩阵 m行n列容易溢出，这点很难把握
57 目前的经验，每次严格按照行-列的顺序进行
58 """
59 class Solution:
60 def longestCommonSubsequence(self, text1, text2):
61 n = len(text1)
62 m = len(text2)
63 dp = [[0]*(m+1) for _ in range(n+1)] #建立 n+1行 m+1列矩阵，值全为0
64 for i in range(1, n+1): #bug 内外循环层写反了，导致溢出,n+1 * m+1 矩阵
65 for j in range(1, m+1):
66 if text1[i-1] == text2[j-1]:
67 dp[i][j] = dp[i-1][j-1] + 1
68 else:
69 dp[i][j] = max(dp[i-1][j], dp[i][j-1])
70 return dp[-1][-1]
