leetcode第十題（java解法）

Given an input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*'.

'.' Matches any single character.
'*' Matches zero or more of the preceding element.

The matching should cover the entire input string (not partial).

Note:

• s could be empty and contains only lowercase letters a-z.
• p could be empty and contains only lowercase letters a-z, and characters like . or *.

Example 1:

Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input:
s = "aa"
p = "a*"
Output: true
Explanation: '*' means zero or more of the precedeng element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

Input:
s = "ab"
p = ".*"
Output: true
Explanation: ".*" means "zero or more (*) of any character (.)".

Example 4:

Input:
s = "aab"
p = "c*a*b"
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab".

Example 5:

Input:
s = "mississippi"
p = "mis*is*p*."
Output: false

 

借鑑了網上博主c語言的思路，從後往前匹配，不用考慮越界情況。(回溯法)

public static boolean isMatch(String s, String p) {
    return myMacth(s,s.length()-1,p,p.length()-1);
}
private static boolean myMacth(String s,int i,String p,int j){
    System.out.println(i+"  "+j);
    if(j==-1)
        if(i==-1)
            return true;
    else return false;
    if(p.charAt(j)=='*'){//若碰到*，則先是判斷*前字符與s中是否相等，相等則下一步，不相等則跳過這兩位，進入下一步。
        if(i >= 0 &&(p.charAt(j-1)=='.'||p.charAt(j-1)==s.charAt(i))){
            if(myMacth(s,i-1,p,j))
                return true;
        }
        return myMacth(s,i,p,j-2);
    }
    if(i==-1&&j>=0) return false;//此處判斷很關鍵，s先遍歷完的情況下，此時p沒有遍歷完。
    if(p.charAt(j)=='.'||p.charAt(j)==s.charAt(i))
        return myMacth(s,i-1,p,j-1);
    return false;
}