41. 缺失的第一个正数

class Solution {
public:
    int firstMissingPositive(vector<int>& nums) {
        int len = nums.size();
        for (int i = 0; i < len; ++ i) {
            while (nums[i] > 0 && nums[i] <= len && nums[nums[i] - 1] != nums[i])
                swap(nums[i], nums[nums[i] - 1]);
        }
        for (int i = 0; i < len; ++ i)
            if (nums[i] != i + 1) return i + 1;
        return len + 1;
    }
};

42. 接雨水

class Solution {
public:
    int trap(vector<int>& height) {
        stack<int> st;
        int ans = 0;
        for (int i = 0; i < height.size(); ++ i) {
            while (st.size() && height[st.top()] < height[i]) {
                int x = st.top();st.pop();
                if (!st.size()) break;
                int l = st.top();
                int h = min(height[i], height[l]) - height[x];
                int w = i - l - 1;
                ans += h * w;
            }
            st.push(i);
        }
        return ans;
    }
};

43. 字符串相乘

class Solution {
public:
    string multiply(string num1, string num2) {
        if (num1 == "0" || num2 == "0") return "0";
        vector<int> ve(num1.size() + num2.size());
        for (int i = num1.size() - 1; i >= 0; -- i)
        for (int j = num2.size() - 1; j >= 0; -- j) {
            int val = ve[i + j + 1] + (num1[i] - '0') * (num2[j] - '0');
            ve[i + j] += val / 10;
            ve[i + j + 1] = val % 10;
        }
        string ans;
        for (int i = 0; i < ve.size(); ++ i) {
            if (!i && !ve[i]) continue;
            ans.push_back(ve[i] + '0');
        }
        return ans;
    }
};

44. 通配符匹配

class Solution {
public:
    bool isMatch(string s, string p) {
        int n = s.size(), m = p.size();
        vector<vector<int>> f(n + 1, vector<int>(m + 1));
        f[0][0] = 1;
        for (int i = 1; i <= m; ++ i)
            if (p[i - 1] == '*') f[0][i] = 1;
            else break;
        for (int i = 1; i <= n; ++ i)
        for (int j = 1; j <= m; ++ j) {
            switch(p[j - 1]) {
                case '?': f[i][j] = f[i - 1][j - 1]; break;
                case '*': f[i][j] = f[i - 1][j] || f[i][j - 1] || f[i - 1][j - 1]; break;
                default: f[i][j] = (s[i - 1] == p[j - 1]) && f[i - 1][j - 1];
            }
        }
        return f[n][m];
    }
};

45. 跳跃游戏 II

class Solution {
public:
    int jump(vector<int>& nums) {
        int n = nums.size() - 1;
        vector<int> f(n + 1, 0x3f3f3f3f);
        f[0] = 0;
        for (int i = 0; i <= n; ++ i) {
            for (int j = 1; j <= nums[i] && i + j <= n; ++ j)
                f[i + j] = min (f[i + j], f[i] + 1);
        }
        return f[n];
    }
};

46. 全排列

class Solution {
public:
    void dfs (int cur, vector<int>& vis, vector<int>& temp, vector<int>& num, vector<vector<int>>& ans) {
        if (cur > num.size()) {
            ans.push_back(temp);
            return;
        }
        for (int i = 1; i <= num.size(); ++ i) {
            if (vis[i - 1]) continue;
            vis[i - 1] = 1;
            temp.push_back(num[i - 1]);
            dfs (cur + 1, vis, temp, num, ans);
            temp.pop_back();
            vis[i - 1] = 0;
        }
    }
    vector<vector<int>> permute(vector<int>& nums) {
        vector<vector<int>> ans;
        vector<int> vis(nums.size(), 0);
        vector<int> temp;
        dfs (1, vis, temp, nums, ans);
        return ans;
    }
};

47. 全排列 II

class Solution {
public:
    void dfs (int cur, vector<int>& vis, vector<int>& temp, vector<int>& num, vector<vector<int>>& ans) {
            if (cur > num.size()) {
                ans.push_back(temp);
                return;
            }
            for (int i = 1; i <= num.size(); ++ i) {
                if (vis[i - 1]) continue;
                if (i != 1 && num[i - 1] == num[i - 2] && !vis[i - 2]) {
                    continue;
                }
                vis[i - 1] = 1;
                temp.push_back(num[i - 1]);
                dfs (cur + 1, vis, temp, num, ans);
                temp.pop_back();
                vis[i - 1] = 0;
            }
        }
    vector<vector<int>> permuteUnique(vector<int>& nums) {
        vector<vector<int>> ans;
        vector<int> vis(nums.size(), 0);
        vector<int> temp;
        sort(nums.begin(), nums.end());
        dfs (1, vis, temp, nums, ans);
        return ans;
    }
};

48. 旋转图像

class Solution {
public:
    void dfs (int x1, int x2, vector<vector<int>>& m) {
        if (x2 - x1 < 1) return;
        int n = m.size() - 1;
        for (int i = x1; i < x2; ++ i) {
            int x = x1, y = i;
            int temp = m[x][y];
            m[x][y] = m[n - y][x];
            m[n - y][x] = m[n - x][n - y];
            m[n - x][n - y] = m[y][n - x];
            m[y][n - x] = temp;
        }
        dfs (x1 + 1, x2 - 1, m);
    }
    void rotate(vector<vector<int>>& matrix) {
        dfs (0, matrix.size() - 1, matrix);
    }
};

49. 字母异位词分组

class Solution {
public:
    vector<vector<string>> groupAnagrams(vector<string>& strs) {
        map<string, vector<string>> mp;
        for (auto& i: strs) {
            string t = i;
            sort(t.begin(), t.end());
            mp[t].push_back(i);
        }
        vector<vector<string>> ans;
        for (auto& [x, y]: mp) {
            ans.push_back(y);
        }
        return ans;
    }
};

50. Pow(x, n)

class Solution {
public:
    double myPow(double x, int n) {
        long f = n < 0 ? -1 : 1, y = abs(n);
        double ans = 1, t = x;
        if (f == -1) t = 1 / t;
        while (y) {
            if (y & 1) ans *= t;
            t *= t;
            y >>= 1;
        }
        return ans;
    }
};