请将LeetCode的前10题完整展现出来，并且使用C加加语言写出最优解法的代码。。 代码请使用c加加语法格式

LeetCode前10题的完整列表如下：

1. 两数之和
2. 两数相加
3. 无重复字符的最长子串
4. 寻找两个有序数组的中位数
5. 最长回文子串
6. Z 字形变换
7. 整数反转
8. 字符串转换整数 (atoi)
9. 回文数
10. 正则表达式匹配

以下是每道题的最优解法的C++代码：

1. 两数之和

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        unordered_map<int, int> numToIndex;
        vector<int> result;

        for (int i = 0; i < nums.size(); i++) {
            int complement = target - nums[i];
            if (numToIndex.count(complement)) {
                result.push_back(numToIndex[complement]);
                result.push_back(i);
                return result;
            }
            numToIndex[nums[i]] = i;
        }

        return result;
    }
};

2. 两数相加

class Solution {
public:
    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
        ListNode* dummyHead = new ListNode(0);
        ListNode* curr = dummyHead;
        int carry = 0;

        while (l1 != nullptr || l2 != nullptr) {
            int x = (l1 != nullptr) ? l1->val : 0;
            int y = (l2 != nullptr) ? l2->val : 0;
            int sum = carry + x + y;
            carry = sum / 10;
            curr->next = new ListNode(sum % 10);
            curr = curr->next;
            if (l1 != nullptr) l1 = l1->next;
            if (l2 != nullptr) l2 = l2->next;
        }

        if (carry > 0) {
            curr->next = new ListNode(carry);
        }

        ListNode* result = dummyHead->next;
        delete dummyHead;
        return result;
    }
};

3. 无重复字符的最长子串

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        unordered_set<char> charSet;
        int left = 0, right = 0, maxLength = 0;

        while (right < s.length()) {
            if (charSet.count(s[right]) == 0) {
                charSet.insert(s[right]);
                maxLength = max(maxLength, right - left + 1);
                right++;
            } else {
                charSet.erase(s[left]);
                left++;
            }
        }

        return maxLength;
    }
};

4. 寻找两个有序数组的中位数

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        int m = nums1.size(), n = nums2.size();
        if (m > n) {
            return findMedianSortedArrays(nums2, nums1);
        }

        int iMin = 0, iMax = m, halfLen = (m + n + 1) / 2;
        while (iMin <= iMax) {
            int i = (iMin + iMax) / 2;
            int j = halfLen - i;
            if (i < iMax && nums2[j - 1] > nums1[i]) {
                iMin = i + 1;
            } else if (i > iMin && nums1[i - 1] > nums2[j]) {
                iMax = i - 1;
            } else {
                int maxLeft = 0;
                if (i == 0) {
                    maxLeft = nums2[j - 1];
                } else if (j == 0) {
                    maxLeft = nums1[i - 1];
                } else {
                    maxLeft = max(nums1[i - 1], nums2[j - 1]);
                }
                if ((m + n) % 2 == 1) {
                    return maxLeft;
                }

                int minRight = 0;
                if (i == m) {
                    minRight = nums2[j];
                } else if (j == n) {
                    minRight = nums1[i];
                } else {
                    minRight = min(nums1[i], nums2[j]);
                }

                return (maxLeft + minRight) / 2.0;
            }
        }
        return 0.0;
    }
};

5. 最长回文子串

class Solution {
public:
    string longestPalindrome(string s) {
        int n = s.length();
        int maxLength = 1, start = 0;
        vector<vector<int>> dp(n, vector<int>(n, 0));

        for (int i = 0; i < n; i++) {
            dp[i][i] = 1;
            if (i < n - 1 && s[i] == s[i + 1]) {
                dp[i][i + 1] = 1;
                maxLength = 2;
                start = i;
            }
        }

        for (int len = 3; len <= n; len++) {
            for (int i = 0; i + len - 1 < n; i++) {
                int j = i + len - 1;
                if (s[i] == s[j] && dp[i + 1][j - 1] == 1) {
                    dp[i][j] = 1;
                    maxLength = len;
                    start = i;
                }
            }
        }

        return s.substr(start, maxLength);
    }
};

6. Z 字形变换

class Solution {
public:
    string convert(string s, int numRows) {
        if (numRows == 1) {
            return s;
        }

        vector<string> rows(min(numRows, int(s.length())));
        int curRow = 0;
        bool goingDown = false;

        for (char c : s) {
            rows[curRow] += c;
            if (curRow == 0 || curRow == numRows - 1) {
                goingDown = !goingDown;
            }
            curRow += goingDown ? 1 : -1;
        }

        string result;
        for (string row : rows) {
            result += row;
        }
        return result;
    }
};

7. 整数反转

class Solution {
public:
    int reverse(int x) {
        int result = 0;

        while (x != 0) {
            int pop = x % 10;
            x /= 10;
            if (result > INT_MAX / 10 || (result == INT_MAX / 10 && pop > 7)) {
                return 0;
            }
            if (result < INT_MIN / 10 || (result == INT_MIN / 10 && pop < -8)) {
                return 0;
            }
            result = result * 10 + pop;
        }

        return result;
    }
};

8. 字符串转换整数 (atoi)

class Solution {
public:
    int myAtoi(string s) {
        int i = 0, sign = 1, result = 0;

        while (i < s.length() && s[i] == ' ') {
            i++;
        }

        if (i < s.length() && (s[i] == '+' || s[i] == '-')) {
            sign = (s[i++] == '-') ? -1 : 1; 
        }

        while (i < s.length() && isdigit(s[i])) {
            int digit = s[i++] - '0';
            if (result > INT_MAX / 10 || (result == INT_MAX / 10 && digit > 7)) {
                return (sign == 1) ? INT_MAX : INT_MIN;
            }
            result = result * 10 + digit;
        }

        return result * sign;
    }
};

9. 回文数

class Solution {
public:
    bool isPalindrome(int x) {
        if (x < 0 || (x % 10 == 0 && x != 0)) {
            return false;
        }

        int reverse = 0;
        while (x > reverse) {
            reverse = reverse * 10 + x % 10;
            x /= 10;
        }

        return x == reverse || x == reverse / 10;
    }
};

10. 正则表达式匹配

class Solution {
public:
    bool isMatch(string s, string p) {
        int m = s.length(), n = p.length();
        vector<vector<bool>> dp(m + 1, vector<bool>(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 && (s[i - 1] == p[j - 1] || p[j - 1] == '.') && dp[i - 1][j - 1];
                }
            }
        }

        return dp[m][n];
    }
};