LeetCode 솔루션 분류
[12/15] 1143. Longest Common Subsequence
본문
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 <= 1000text1andtext2consist of only lowercase English characters.
Accepted
614.7K
Submissions
<div class="text-label-1 dark:text-dark-label-1 text-sm font-medium" style="border: 0px solid; box-sizing: border-box; --tw-border-spacing-x:0; --tw-border-spacing-y:0; --tw-translate-x:0; --tw-translate-y:0; --tw-rotate:0; --tw-skew-x:0; --tw-skew-y:0; --tw-scale-x:1; --tw-scale-y:1; --tw-pan-x: ; --tw-pan-y: ; --tw-pinch-zoom: ; --tw-scroll-snap-strictness:proximity; --tw-ordinal: ; --tw-slashed-zero: ; --tw-numeric-figure: ; --tw-numeric-spacing: ; --tw-numeric-fraction: ; --tw-ring-inset: ; --tw-ring-offset-width:0px; --tw-ring-offset-color:#fff; --tw-ring-color:rgba(59,130,246,0.5); --tw-ring-offset-shadow:0 0 #0000; --tw-ring-shadow:0 0 #0000; --tw-shadow:0 0 #0000; --tw-shadow-colored:0 0 관련자료
-
링크
댓글 1
학부유학생님의 댓글
- 익명
- 작성일
class Solution:
def longestCommonSubsequence(self, text1: str, text2: str) -> int:
# a f e
# 0 0 0 0
# a 0 1 1 1
# b 0 1 1 1
# c 0 1 1 1
# a 0 1 1 1
# e 0 1 1 2
dp = [ [ 0 for _ in range(len(text1)+1) ] for _ in range(len(text2)+1) ]
for i in range(len(text2)):
for j in range(len(text1)):
if text1[j] == text2[i]:
dp[i+1][j+1] = dp[i][j]+1
else:
dp[i+1][j+1] = max(dp[i+1][j], dp[i][j+1])
return dp[-1][-1]
