【dp】leetcode 115. Distinct Subsequences
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【dp】leetcode 115. Distinct Subsequences
dp 与公共子串匹配的算法思想类似
当当前字符s与t匹配时,次数dp[i][j] 可以采用两种方式进行计算,1)s[j]与t[i]匹配 为 dp[i-1][j-1] 2) 不考虑新加入的字符 s[j] ,为dp[i][j-1]. 两种情况相加就可以得到了
public class Solution { public int numDistinct(String s, String t) { int[][] dp = new int[t.length()+1][s.length()+1]; for(int j=0;j<=s.length();j++){ dp[0][j] = 1; } for(int i=1;i<=t.length();i++){ for(int j=1;j<=s.length();j++){ if(s.charAt(j-1)==t.charAt(i-1)){ dp[i][j] = dp[i-1][j-1] + dp[i][j-1]; }else{ dp[i][j] = dp[i][j-1]; } } } return dp[t.length()][s.length()]; }}
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