Description
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.
Examples
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.
思路
dp，构造一个2维矩阵，dp[i][j]表示 text1.substring(0, i + 1)和 text2.substring(0, j + 1)有几个相同的元素，这样更新dp[i][j]的状态转移公式就是
- 如果
text1.charAt(i) == text1.charAt(j)——dp[i][j] = dp[i-1][j-1] + 1 - 第二步，执行
dp[i][j] = Math.max(dp[i-1][j], dp[i][j-1], dp[i][j])
代码
class Solution {
public int longestCommonSubsequence(String text1, String text2) {
int[][] dp = new int[text1.length() + 1][text2.length() + 1];
for (int i = 0; i < text1.length(); i++) {
for (int j = 0; j < text2.length(); j++) {
int add = text1.charAt(i) == text2.charAt(j)? 1: 0;
dp[i + 1][j + 1] = dp[i][j] + add;
dp[i + 1][j + 1] = Math.max(dp[i][j + 1], dp[i + 1][j + 1]);
dp[i + 1][j + 1] = Math.max(dp[i + 1][j], dp[i + 1][j + 1]);
}
}
return dp[text1.length()][text2.length()];
}
}
