Simulation volume leetcode [general] 1143 Longest common subsequence

1143. Longest common subsequence

Given two strings  text1 and  text2, Returns the longest of these two strings Common subsequence The length of . If it doesn't exist Common subsequence , return 0 .

A string of   Subsequence   This is a new string ： It is to delete some characters from the original string without changing the relative order of the characters （ You can also delete no characters ） After the composition of the new string .

for example ,“ace” yes “abcde” The subsequence , but “aec” No “abcde” The subsequence .
Two strings of Common subsequence Is the subsequence shared by these two strings .

Example 1：

Input ：text1 = “abcde”, text2 = “ace”
Output ：3
explain ： The longest common subsequence is “ace” , Its length is 3 .
Example 2：

Input ：text1 = “abc”, text2 = “abc”
Output ：3
explain ： The longest common subsequence is “abc” , Its length is 3 .
Example 3：

Input ：text1 = “abc”, text2 = “def”
Output ：0
explain ： The two strings have no common subsequence , return 0 .

Tips ：

1 <= text1.length, text2.length <= 1000
text1 and  text2 It only consists of lowercase English characters .

source ： Power button （LeetCode）
link ：https://leetcode-cn.com/problems/longest-common-subsequence
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Code ：

from leetcode_python.utils import *


class Solution:
    def __init__(self):
        pass

    def longestCommonSubsequence(self, text1: str, text2: str) -> int:
        length1,length2 = len(text1),len(text2)
        if 0==length1 and 0==length2:return 0
        this_dp = [0]*(length1+1)
        for to_word2 in range(length2+1):
            last_dp = this_dp.copy()
            for from_word1 in range(length1+1):
                if 0==from_word1 or 0==to_word2:
                    this_dp[from_word1] = 0
                elif text1[from_word1-1]==text2[to_word2-1]:
                    this_dp[from_word1] = last_dp[from_word1-1] + 1
                else:
                    this_dp[from_word1] = max(this_dp[from_word1 - 1], last_dp[from_word1])
            print(f'to w2[{
      to_word2}-1], {
      this_dp}')
        return this_dp[-1]


    def minDistance_583_ Deletion of two strings (self, word1: str, word2: str) -> int:
        length1,length2 = len(word1),len(word2)
        if 0==length1 and 0==length2:return 0
        this_dp = [0]*(length1+1) # word1  Each letter of to  word2  The number of times each letter needs to be deleted 
        for to_word2 in range(length2+1):
            last_dp = this_dp.copy()
            for from_word1 in range(length1+1):
                if 0==from_word1 or 0==to_word2:   #  consider word1 Delete to 0 The situation of 
                    this_dp[from_word1] = from_word1 + to_word2
                elif word1[from_word1-1]==word2[to_word2-1]:
                    this_dp[from_word1] = last_dp[from_word1-1]
                else:
                    this_dp[from_word1] = min(this_dp[from_word1 - 1], last_dp[from_word1]) + 1
            print(f'to w2[{
      to_word2}-1], {
      this_dp}')
        return this_dp[-1]

def test(data_test):
    s = Solution()
    return s.longestCommonSubsequence(*data_test)


def test_obj(data_test):
    result = [None]
    obj = Solution(*data_test[1][0])
    for fun, data in zip(data_test[0][1::], data_test[1][1::]):
        if data:
            res = obj.__getattribute__(fun)(*data)
        else:
            res = obj.__getattribute__(fun)()
        result.append(res)
    return result


if __name__ == '__main__':
    datas = [
        ["abcde","ace"],
        ["abc","abc"],
        ["abc","def"],
    ]
    for data_test in datas:
        t0 = time.time()
        print('-' * 50)
        print('input:', data_test)
        print('output:', test(data_test))
        print(f'use time:{
      time.time() - t0}s')

remarks ：
GitHub：https://github.com/monijuan/leetcode_python

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