LeetCode 1143 Longest Common Subsequence

LeetCode

Given two strings text1 and text2, return the length of their longest common subsequence. If there is no common subsequence, return 0.

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.

For example, "ace" is a subsequence of "abcde".

A common subsequence of two strings is a subsequence that is common to both strings.

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, text2.length <= 1000
* text1 and text2 consist of only lowercase English characters.

题目大意：

求两字符串的最大公共字符序列，不一定需要连续

解题思路：

两字符串最值问题用DP
dp[i][j]为最大公共字符序列，最后一位不需要相等。递归式为:

dp[i][j] = dp[i - 1][j - 1] + 1 if text1[i - 1] == text2[j - 1]
         = max(dp[i - 1][j], dp[i][j - 1])

解题步骤：

DP五点注意事项

注意事项：

1. 不相等时候不需要dp[i - 1][j - 1]，因为已经包含在dp[i - 1][j]或dp[i][j - 1]中, DP属于累计DP

Python代码：

# dp[i][j] = dp[i - 1][j - 1] + 1 if text1[i - 1] == text2[j - 1]
#          = max(dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - 1])
def longestCommonSubsequence(self, text1: str, text2: str) -> int:
	dp = [[0 for _ in range(len(text2) + 1)] for _ in range(len(text1) + 1)]
	for i in range(1, len(dp)):
		for j in range(1, len(dp[0])):
			if text1[i - 1] == text2[j - 1]:
				dp[i][j] = dp[i - 1][j - 1] + 1
			else:
				dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]) # no dp[i - 1][j - 1] but no impact
	return dp[-1][-1]

打印路径

Python代码：

longest, res = dp[-1][-1], ''
while m >= 0 and n >= 0:
	if dp[m - 1][n] == longest:
		m -= 1
	elif dp[m][n - 1] == longest:
		n -= 1
	else:
		res += text1[m - 1]
		longest -= 1
		m -= 1
		n -= 1
return res[::-1]

算法分析：

时间复杂度为O(nm)，空间复杂度O(nm)