● 392.判断子序列

思路

递推顺序：从上到下 从左到右

代码：

class Solution {public boolean isSubsequence(String s, String t) {int[][]dp=new int[s.length()+1][t.length()+1];for(int i=1;i<=s.length();i++){for(int j=1;j<=t.length();j++){if(s.charAt(i-1)==t.charAt(j-1)){dp[i][j]=dp[i-1][j-1]+1;}else{dp[i][j]=dp[i][j-1];}}}// for(int i=1;i<=s.length();i++){//     for(int j=1;j<=t.length();j++){//         System.out.print(dp[i][j]);//     }//     System.out.println();// }return dp[s.length()][t.length()]==s.length();}
}

● 115.不同的子序列

思路：

dp[i][j]：以i-1为结尾的s子序列中出现以j-1为结尾的t的个数为dp[i][j]。

代码：

class Solution {public int numDistinct(String s, String t) {int[][] dp = new int[s.length() + 1][t.length() + 1];// dp[i][j]：以i-1为结尾的s子序列中出现以j-1为结尾的t的个数为dp[i][j]。for (int i = 0; i < s.length() + 1; i++) {dp[i][0] = 1;}for (int i = 1; i < s.length() + 1; i++) {for (int j = 1; j < t.length() + 1; j++) {if(s.charAt(i-1)==t.charAt(j-1)){dp[i][j]=dp[i-1][j-1]+dp[i-1][j];}else{dp[i][j]=dp[i-1][j];}}}return dp[s.length()][t.length()];}
}