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CyC2018 2018-03-19 15:33:57 +08:00
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notes/剑指 offer 题解.md
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@ -878,9 +878,9 @@ p.charAt(j) == '.' : dp[i][j] = dp[i-1][j-1];
p.charAt(j) == '*' : p.charAt(j) == '*' :
p.charAt(j-1) != s.charAt(i) : dp[i][j] = dp[i][j-2] // in this case, a* only counts as empty p.charAt(j-1) != s.charAt(i) : dp[i][j] = dp[i][j-2] // in this case, a* only counts as empty
p.charAt(j-1) == s.charAt(i) or p.charAt(i-1) == '.': p.charAt(j-1) == s.charAt(i) or p.charAt(i-1) == '.':
dp[i][j] = dp[i-1][j] // in this case, a* counts as multiple a dp[i][j] = dp[i-1][j] // in this case, a* counts as multiple a
or dp[i][j] = dp[i][j-1] // in this case, a* counts as single a or dp[i][j] = dp[i][j-1] // in this case, a* counts as single a
or dp[i][j] = dp[i][j-2] // in this case, a* counts as empty or dp[i][j] = dp[i][j-2] // in this case, a* counts as empty
``` ```
```java ```java