A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not).
Here is an example:
S = "rabbbit", T = "rabbit"
Return 3.
public class Solution { public int numDistinct(String s, String t) { if (s == null) return 0; if (t == null || t.length() == 0) return 1; if (s.length() == 0) return 0; int lenS = s.length(); int lenT = t.length(); int[][] dp = new int[2][lenT + 1]; dp[0][0] = 1; int lastLenS = 0; int currLenS = 1; for (int i = 1; i <= lenS; i++) { currLenS = 1 - lastLenS; dp[currLenS][0] = 1; for (int j = 1; j <= lenT; j++) { dp[currLenS][j] = dp[lastLenS][j]; if (s.charAt(i - 1) == t.charAt(j - 1)) dp[currLenS][j] += dp[lastLenS][j-1]; } lastLenS = currLenS; } return dp[currLenS][lenT]; } // dp[i][j] = dp[i - 1][j] + (s[i - 1] == t[j - 1] ? dp[i - 1][j - 1] : 0); // dp[i][j]表示s的前i字符构成的子串能匹配t中前j个字符构成的子串的子序列数 // 不管s和t当前考察位置处的字符是否相等,dp[i][j]至少有dp[i-1][j] // 如果当前考察位置处两字符相等,则再加上dp[i-1][j-1]的数目 public int numDistinct1(String s, String t) { if (s == null) return 0; if (t == null || t.length() == 0) return 1; if (s.length() == 0) return 0; int lenS = s.length(); int lenT = t.length(); int[][] dp = new int[lenS + 1][lenT + 1]; dp[0][0] = 1; for (int i = 1; i <= lenS; i++) { dp[i][0] = 1; for (int j = 1; j <= lenT; j++) { dp[i][j] = dp[i - 1][j]; if (s.charAt(i - 1) == t.charAt(j - 1)) dp[i][j] += dp[i - 1][j - 1]; } } return dp[lenS][lenT]; } }