Title description
Given two strings text1 and text2, return the length of the longest common subsequence of these two strings.
A subsequence of a character string refers to a new character string: it is a new character string formed by deleting some characters (or no characters) from the original string without changing the relative order of the characters.
For example ,
"ace" 是 "abcde" 的子序列，但 "aec" 不是
"abcde" 的子序列。两个字符串的「公共子序列」
是这两个字符串所共同拥有的子序列。
If the two strings have no common subsequence, 0 is returned.
Example 1:
输入：text1 = "abcde", text2 = "ace"
输出：3
解释：最长公共子序列是 "ace"，它的长度为 3。
Example 2:
输入：text1 = "abc", text2 = "abc"
输出：3
解释：最长公共子序列是 "abc"，它的长度为 3。
Example 3:
输入：text1 = "abc", text2 = "def"
输出：0
解释：两个字符串没有公共子序列，返回 0。
prompt:
1 <= text1.length <= 1000
1 <= text2.length <= 1000
输入的字符串只含有小写英文字符。
Source: LeetCode (LeetCode)
Link: https://leetcode-cn.com/problems/longest-common-subsequence
Copyright is owned by LeetCode . For commercial reprints, please contact the official authorization. For non-commercial reprints, please indicate the source.
Problem-solving ideas:
How about DP? What do we set for the title. Since it is two strings, we will set up a two-dimensional array dp[i][j], which represents the first i characters of the first character and the second The longest subsequence that can be formed by the first j characters of the characters. The example shows that the subsequence can be discontinuous, so the longest affirmation dp[t1.length][t2.length].
Then we look at the last step,
1. What are if t1.charAt(t1.length-1)和t2.charAt(t2.length-1)不相等;
we going to do? Is directly equal to dp[t1.length-1-1][t2.length-1-1]? No, because this subsequence can be discontinuous, let's draw a picture:
Say yes 不包含两者中的至少一个, why not include dp[i-1][j-1]? It is possible to add it to the state transition equation, and the correct answer can be obtained, but it is not necessary, because dp[i-1][j-1] must be 小于等于dp[i][j-1], dp[i-1] [j], max is impossible to get it. The larger i and j, the more the same palindrome that it may contain. It will be clear if you draw it on paper. I am lazy to not draw it hahahaha
2. In the case of equality, directly dp[i][j]=dp[i-1][j-1]+1; this should be well understood
and then code:
class Solution {
public int longestCommonSubsequence(String text1, String text2) {
char s1[]=text1.toCharArray();
char s2[]=text2.toCharArray();
int x=s1.length;
int y=s2.length;
int dp[][]=new int[x+1][y+1];
for(int i=1;i<=x;i++){
for(int j=1;j<=y;j++){
if(s1[i-1]==s2[j-1])
dp[i][j]=dp[i-1][j-1]+1;
else{
dp[i][j]=Math.max(dp[i-1][j],dp[i][j-1]);
}
}
}
return dp[x][y];
}
}