Longest Common Subsequence

最长公共子序列, 这题就是dp.

dp[i][j]表示在第一个string的i位和第二个string的j位的最长公共子序列.

dp[i][j] = dp[i – 1][j – 1] when A[i] == B[j], otherwise dp[i][j] = max(dp[i][j – 1], dp[i – 1][j])

class Solution {
    public int longestCommonSubsequence(String text1, String text2) {
        int n = text1.length();
        int m = text2.length();
        int[][] dp = new int[n][m];
        for(int i = 0; i < n; i++) {
            if(text1.charAt(i) == text2.charAt(0))
                dp[i][0] = 1;
            else if(i > 0 && dp[i - 1][0] == 1)
                dp[i][0] = 1;
        }
        for(int j = 0; j < m; j++) {
            if(text1.charAt(0) == text2.charAt(j))
                dp[0][j] = 1;
            else if(j > 0 && dp[0][j - 1] == 1)
                dp[0][j] = 1;
                
        }
        for(int i = 1; i < n; i++){
            for(int j = 1; j < m; j++){
                if(text1.charAt(i) == text2.charAt(j))
                    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[n - 1][m - 1];
    }
}