Time: 20190906
Type: Medium
给定两个字符串text1
和text2
,返回它们的最长公共子序列。
子序列是通过原字符串删除一些字符(可以不删)生成的新的字符串,不改变原来字符的相对顺序。比如ace
是abcde
的子序列,而aec
则不是。
公共子序列是指该序列是两个字符串都有的子序列。
如果没有公共子序列,则返回0即可。
Example 1:
Input: text1 = “abcde”, text2 = “ace”
Output: 3
Explanation: The longest common subsequence is “ace” and its length is 3.
Example 2:
Input: text1 = “abc”, text2 = “abc”
Output: 3
Explanation: The longest common subsequence is “abc” and its length is 3.
Example 3:
Input: text1 = “abc”, text2 = “def”
Output: 0
Explanation: There is no such common subsequence, so the result is 0.
Constraints:
1 <= text1.length <= 1000
1 <= text2.length <= 1000
The input strings consist of lowercase English characters only.
本题是动态规划经典题型了。
两个字符串,因此可以想到用二维DP来表达状态。
f[i][j]
表示text1
中到i
下标为止的字符和text2
中到j
为止的下标的字符的最长公共子序列长度。
f[i][j]
的求解可以向左查看三个状态:
f[i-1][j-1]
f[i][j-1]
f[i-1][j]
如果当前字符相等,则:f[i][j] = f[i-1][j-1] + 1
,否则:f[i][j] = max(f[i-1][j], f[i][j-1])
。
class Solution:
def longestCommonSubsequence(self, text1: str, text2: str) -> int:
m = len(text1)
n = len(text2)
if m < 1 or n < 1:
return 0
f = [[0] * (n + 1) for _ in range(m + 1) ]
# f[i][j]表示text1中[1...i], text2中[1...j]LCS的长度
# f[i][j] = f[i-1][j-1] if text1[i-1] != text2[j-1] else max(f[i-1][j], f[i][j-1])
res = ""
for i in range(1, m + 1):
for j in range(1, n + 1):
if text1[i-1] == text2[j-1]:
res += text1[i-1] #
f[i][j] = f[i-1][j-1] + 1
else:
f[i][j] = max(f[i-1][j], f[i][j-1])
print(res) # 即为最长公共子序列
return f[m][n]
本题有个小技巧是从i=1
和j=1
开始使用,如果从0开始就比较难处理f[1][0]和f[0][1]两种状态。因为从0开始,f[1][0]
表示text1
提供两个字符,text2
提供1个字符,需要向左再开拓一个状态表示一个提供1个字符,一个不提供字符的情况。
class Solution:
def longestCommonSubsequence(self, text1: str, text2: str) -> int:
m = len(text1)
n = len(text2)
if m < 1 or n < 1:
return 0
f = [[0] * (n + 1) for _ in range(m + 1) ]
# f[i][j]表示text1中[1...i], text2中[1...j]LCS的长度
# f[i][j] = f[i-1][j-1] if text1[i-1] != text2[j-1] else max(f[i-1][j], f[i][j-1])
for i in range(1, m + 1):
for j in range(1, n + 1):
if text1[i-1] == text2[j-1]:
f[i][j] = f[i-1][j-1] + 1
else:
f[i][j] = max(f[i-1][j], f[i][j-1])
# Following code is used to print LCS
index = f[m][n]
# Create a character array to store the lcs string
lcs = [""] * (index + 1)
lcs[index] = ""
# Start from the right-most-bottom-most corner and
# one by one store characters in lcs[]
i = m
j = n
while i > 0 and j > 0:
# If current character in X[] and Y are same, then
# current character is part of LCS
if text1[i-1] == text2[j-1]:
lcs[index-1] = text1[i-1]
i -= 1
j -= 1
index -= 1
# If not same, then find the larger of two and
# go in the direction of larger value
elif f[i-1][j] > f[i][j-1]:
i -= 1
else:
j -= 1
print("lcs is: ", lcs) # lcs是数组形式
return f[m][n]
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END.