【LeetCode010算法/编程练习C++】Regular Expression Matching//挺烦的一条……

10. Regular Expression Matching

• Total Accepted: 113015 
• Total Submissions: 481908 
• Difficulty: Hard 
• Contributors: Admin

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).

The function prototype should be:
bool isMatch(const char *s, const char *p)

Some examples:
isMatch("aa","a") → false
isMatch("aa","aa") → true
isMatch("aaa","aa") → false
isMatch("aa", "a*") → true
isMatch("aa", ".*") → true
isMatch("ab", ".*") → true
isMatch("aab", "c*a*b") → true

感觉这题最关键的是把题目意思理解了……

1.首先需要注意的是：匹配的意思是，整个都一样！！互相能转化

2.“.*”匹配任意

3.X*的意思是有任意个包括0个X

4..可以替代任意一个字符

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先来一个最烂的但是能解决的方法：//调试了很多次……

思路：不断迭代，通过调试剔除了很多种情况……

class Solution {
public:
	bool isMatch(string s, string p) {
	string temp = "";
	bool result=false;
	if (s == "") {
		if (p.size()>1 && p[1] == '*')result = isMatch(s, p.substr(2, p.size() - 2));
	}
	if (s == p||(p.size()==1&&p[0]=='.'&&s.size()==1))result = true;
	else if (p == "")result = false;
	else if (p.size() == 1 && s != p)result = false;
	else if (p.size()>1&&s.size()>0) {
		if ((s[0] == p[0]||p[0]=='.') && p[1] != '*')result = isMatch(s.substr(1, s.size() - 1), p.substr(1, p.size() - 1));
		if ((s[0] != p[0] && p[0] != '.') && p[1] == '*')result = isMatch(s, p.substr(2, p.size() - 2));
		if ((s[0] != p[0] && p[0] != '.') && p[1] != '*')result = false;
		if ((s[0] == p[0] || p[0] == '.') && p[1] == '*') {
			if (p[0] == '.') {
				for (int i = 0; i < s.size()+1; i++) {
					bool tempp = false;
					tempp = isMatch(s.substr(i, s.size() - i), p.substr(2, p.size() - 2));
					result += isMatch(s.substr(i, s.size() - i), p.substr(2, p.size() - 2));
				}

			}
			else {
				int n = 1, k = 1, m = 1, i = 1, ii = 0;//p里k个a，m个*。。。s里n个a
				while (n < s.size() && s[n] == s[0]) {
					n++;
				}
				while (m + k < p.size()  && ((p[m + k] == p[0]) || p[m + k] == '*')) {
					if (p[m + k] == p[0])k++;
					else m++;
				}
				if (k - m > n)result = false;
				else {
					for(int a_temp=k-m;a_temp

没错，运行结果击败了0.26%的人……666ms，没runtime limited真的谢天谢地了。

再来一种Top Solution里的解决方法：  9行16ms……//肝脑涂地

class Solution {
public:
  bool isMatch(string s, string p) {
        int m = s.length(), n = p.length(); 
        vector > dp(m + 1, vector (n + 1, false));
        dp[0][0] = true;
        for (int i = 0; i <= m; i++)
            for (int j = 1; j <= n; j++)
                if (p[j - 1] == '*')
                    dp[i][j] = dp[i][j - 2] || (i > 0 && (s[i - 1] == p[j - 2] || p[j - 2] == '.') && dp[i - 1][j]);
                else dp[i][j] = i > 0 && dp[i - 1][j - 1] && (s[i - 1] == p[j - 1] || p[j - 1] == '.');
        return dp[m][n];
    }

};

真的好简洁…………