题目链接392.判断子序列
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
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];
}
}