Sword finger offer II 095 Longest common subsequence

Given two strings text1 and text2, Returns the longest of these two strings Common subsequence The length of . If it doesn't exist Common subsequence , return 0 .

A string of Subsequence This is a new string ： It is to delete some characters from the original string without changing the relative order of the characters （ You can also delete no characters ） After the composition of the new string .

for example ,“ace” yes “abcde” The subsequence , but “aec” No “abcde” The subsequence .
Two strings of Common subsequence Is the subsequence shared by these two strings .

Example 1：

Input ：text1 = “abcde”, text2 = “ace”
Output ：3
explain ： The longest common subsequence is “ace” , Its length is 3 .
Example 2：

Input ：text1 = “abc”, text2 = “abc”
Output ：3
explain ： The longest common subsequence is “abc” , Its length is 3 .
Example 3：

Input ：text1 = “abc”, text2 = “def”
Output ：0
explain ： The two strings have no common subsequence , return 0 .

source ： Power button （LeetCode）
link ：https://leetcode-cn.com/problems/qJnOS7

Dynamic programming ：

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++) {
    
           char c1 = text1.charAt(i - 1);
           for (int j = 1; j <= n; j++) {
    
               char c2 = text2.charAt(j - 1);
               if (c1 == c2) {
    
                   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[m][n];
   }
}

dp[i][j] The meaning of ：text1 Of 0 ~ i String and text2 Of 0 ~ j The common subsequence length of the string of .

First, we can determine the boundary, that is dp[i][0] and dp[0][i] = 0, Because the length is 0 The length of the common subsequence of the string must also be 0.
Because array initialization is 0, So you don't have to initialize the boundary again .

When subscript i And j Situated The letters are the same , explain dp[i][j] = dp[i-1][j-1] + 1,+1 Refer to subscript i and j Can be directly added to the longest common subsequence .

If the letters are different , be Compare dp[i-1][j] and dp[i][j-1] In both cases , Take the larger .

From a two-dimensional perspective , Whether it's dp[i-1][j] still dp[i][j-1] All in dp[i][j] Top left of , So we have i It is an external heavy cycle j Traverse for inner loop , You can finish the whole dp The calculation of array .
Eventually return dp[m][n].