题目链接392.判断子序列

复制代码

class Solution {
    public boolean isSubsequence(String s, String t) {
        int length1 = s.length();
        int length2 = t.length();
        int[][] dp = new int[length1+1][length2+1];
        for(int i = 1; i <= length1; i++){
            for(int j = 1; j <=length2; 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];
                }
            }
        }
        return dp[length1][length2] == length1 ? true:false; 
    }
}

题目链接不同的子序列

复制代码

class Solution {
    public int numDistinct(String s, String t) {
        int length1 = s.length();
        int length2 = t.length();
        int[][] dp = new int[length1+1][length2+1];
        for(int i = 0; i <= length1; i++){
            dp[i][0] = 1;
        }
        for(int i =1; i <= length1; i++){
            for(int j =1; j <=length2; 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[length1][length2];
    }
}