• 算法
    • 动态规划
    • 1.dp[i][j]表示str1[0,i-1]和str2[0,j-1]的最长公共子序列
    • 2.初始化dp[x][0] = 0, dp[0][x] = 0
    • 3.过渡公式
      • 如果str1[i]==str2[j], dp[i][j] = dp[i-1][j-1] + 1;
      • 如果str1[i]!=str2[j], dp[i][j] = MAX(dp[i-1][j], dp[i][j-1]); 
public int longestCommonSubsequence(String text1, String text2) {
    int[][] dp = new int[text1.length()+1][text2.length()+1];
    for (int i = 1; i <= text1.length(); i++) {
        for (int j = 1; j <= text2.length(); j++) {
            if (text1.charAt(i-1) == text2.charAt(j-1)) {
                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[text1.length()][text2.length()];
}