Link: https://leetcode.cn/problems/longest-common-subsequence/
Question
difficulty: mid
adj diff: 2
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.
DP,不难。
Code
public int longestCommonSubsequence(String text1, String text2) {
int m = text1.length();
int n = text2.length();
int[][] dp = new int[m][n];
char[] s1 = text1.toCharArray();
char[] s2 = text2.toCharArray();
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
// dp[i][j] represent the LCS up till s1[i] and s2[j] inclusive
// if char is same, dp[i][j] = 1 + dp[i-1][j-1]
// else, dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1])
if (i == 0 && j == 0) {
dp[i][j] = s1[i] == s2[j] ? 1 : 0;
} else if (i == 0) {
dp[i][j] = s1[i] == s2[j] ? 1 : dp[i][j-1];
} else if (j == 0) {
dp[i][j] = s1[i] == s2[j] ? 1 : dp[i-1][j];
} else {
dp[i][j] = s1[i] == s2[j] ? 1 + dp[i-1][j-1] : Math.max(dp[i-1][j], dp[i][j-1]);
}
}
}
return dp[m - 1][n - 1];
}