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CyC2018 2018-03-19 15:45:02 +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-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) == '.':
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-2] // in this case, a* counts as empty
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-2] // in this case, a* counts as empty
```
```java
@ -920,8 +920,7 @@ public boolean match(char[] str, char[] pattern) {
```java
public boolean isNumeric(char[] str) {
String string = String.valueOf(str);
return string.matches("[\\+-]?[0-9]*(\\.[0-9]*)?([eE][\\+-]?[0-9]+)?");
return new String(str).matches("[+-]?\\d*(\\.\\d+)?([eE][+-]?\\d+)?");
}
```