原题链接在这里：https://leetcode.com/problems/longest-common-subsequence/
题目：
Given two strings text1 and text2, return the length of their longest common subsequence.
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. (eg, "ace" is a subsequence of "abcde" while "aec" is not). A common subsequence of two strings is a subsequence that is common to both strings.
If there is no common subsequence, return 0.
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 <= 10001 <= text2.length <= 1000- The input strings consist of lowercase English characters only.
题解：
dp[i][j] stands for length of LCS between text1 up to i and text2 up to j.
If text1.charAt(i) == text2.charAt(j), dp[i][j] = dp[i-1][j-1] + 1.
Otherwise, dp[i][j] = Math.max(dp[i][j-1], dp[i-1][j]).
如果不放心的话，就直接取上述三个的最小.
Time Complexity: O(m*n). m = text1.length. n = text2.length.
Space: O(m*n).
AC Java:
1 class Solution { 2 public int longestCommonSubsequence(String text1, String text2) { 3 if(text1 == null || text1.length() == 0 || text2 == null || text2.length() == 0){ 4 return 0; 5 } 6 7 int m = text1.length(); 8 int n = text2.length(); 9 int [][] dp = new int[m+1][n+1]; 10 for(int i = 0; i<m; i++){ 11 for(int j = 0; j<n; j++){ 12 dp[i+1][j+1] = Math.max(dp[i][j+1], dp[i+1][j]); 13 if(text1.charAt(i) == text2.charAt(j)){ 14 dp[i+1][j+1] = Math.max(dp[i+1][j+1], dp[i][j]+1); 15 } 16 } 17 } 18 19 return dp[m][n]; 20 } 21 }
来源：博客园
作者：Dylan_Java_NYC
链接：https://www.cnblogs.com/Dylan-Java-NYC/p/11450819.html