regular expression matching

Implement regular expression matching with support for ‘.’ and ‘*’.

note:

'.' 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

######思路：回溯

class Solution {
public:
    bool judgePoint24(vector& nums) {
        vector cards(nums.size(), 0);
        for (int i = 0; i < nums.size(); i++)
            cards[i] = (double)nums[i];
        return backTracking(cards);
    }
    bool backTracking(vector& cards) {
        if (cards.size() == 0)
            return false;
        if (cards.size() == 1)
            return abs(cards[0] - 24) < 1e-6;

        for (int i = 0; i < cards.size(); i++) {
            for (int j = 0; j < cards.size(); j++) {
                if (i == j)
                    continue;

                vector tempCards;
                for (int k = 0; k < cards.size(); k++)
                    if (k != i && k != j)
                        tempCards.push_back(cards[k]);

                // '+','*','-','/'
                for (int k = 0; k < 4; k++) {
                    if (k < 2 && j < i)
                        continue;
                    if (k == 0) tempCards.push_back(cards[i] + cards[j]);
                    else if (k == 1) tempCards.push_back(cards[i] * cards[j]);
                    else if (k == 2) tempCards.push_back(cards[i] - cards[j]);
                    else if (cards[j] == 0) continue;
                    else tempCards.push_back(cards[i] / cards[j]);

                    if (backTracking(tempCards))
                        return true;

                    tempCards.pop_back();
                }
            }
        }

        return false;
    }
};