1143. Longest Common Subsequence
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" is a subsequence of "abcde", but "aec" is not a subsequence of "abcde". The "common subsequence" of two strings is the subsequence shared by the two strings.If the two strings have no common subsequence, 0 is returned.
The longest common subsequence (Longest Common Subsequence, abbreviated LCS) is a very classic interview question, because its solution is typical 二维动态规划, most of the more difficult string problems are the same as this question, for example 编辑距离. Moreover, this algorithm can be used to solve other problems with a little modification, so the LCS algorithm is worth mastering.
For analysis, see this article " Classic Dynamic Programming: The Longest Common Subsequence "
class Solution {
public int longestCommonSubsequence(String text1, String text2) {
int m = text1.length(), n = text2.length();
int[][] dp = new int[m+1][n+1];
for(int i = 1; i <= m; i++){
for(int j = 1; j <= n; j++){
if(text1.charAt(i-1) == text2.charAt(j-1)){
dp[i][j] = 1 + dp[i-1][j-1];
}else {
dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1]);
}
}
}
return dp[m][n];
}
}
Save time by converting to character array
class Solution {
public int longestCommonSubsequence(String text1, String text2) {
char[] t1 = text1.toCharArray();
char[] t2 = text2.toCharArray();
int m = t1.length;
int n = t2.length;
int[][] dp = new int[m+1][n+1];
for (int i = 1; i <= m; i++) {
for (int j = 1; j <= n; j++) {
if (t1[i-1] == t2[j-1]){
dp[i][j] = 1+ dp[i-1][j-1];
}else {
dp[i][j] = Math.max(dp[i-1][j],dp[i][j-1]);
}
}
}
return dp[m][n];
}
}