301. 刪除無效的括號

Tag: 回溯

class Solution {
public:
    vector<string> res;
    vector<string> removeInvalidParentheses(string s) {
        int lremove = 0;
        int rremove = 0;

        for (char c : s) {
            if (c == '(') {
                lremove++;
            } else if (c == ')') {
                if (lremove == 0) {
                    rremove++;
                } else {
                    lremove--;
                }
            }
        }
        helper(s, 0, lremove, rremove);
        return res;
    }

    void helper(string str, int start, int lremove, int rremove) {
        if (lremove == 0 && rremove == 0) {
            if (isValid(str)) {
                res.push_back(str);
            }
            return;
        }
        for (int i = start; i < str.size(); i++) {
            if (i != start && str[i] == str[i - 1]) {
                continue;
            }
            // 如果剩餘的字元無法滿足去掉的數量要求，直接返回
            if (lremove + rremove > str.size() - i) {
                return;
            }
            // 嘗試去掉一個左括號
            if (lremove > 0 && str[i] == '(') {
                helper(str.substr(0, i) + str.substr(i + 1), i, lremove - 1, rremove);
            }
            // 嘗試去掉一個右括號
            if (rremove > 0 && str[i] == ')') {
                helper(str.substr(0, i) + str.substr(i + 1), i, lremove, rremove - 1);
            }
        }
    }

    inline bool isValid(const string& str) {
        int cnt = 0;

        for (int i = 0; i < str.size(); i++) {
            if (str[i] == '(') {
                cnt++;
            } else if (str[i] == ')') {
                cnt--;
                if (cnt < 0) {
                    return false;
                }
            }
        }

        return cnt == 0;
    }
};
 

309. 最佳買賣股票時機含冷凍期

Tag: 動態規劃

class Solution {
public:
    int maxProfit(vector<int>& prices) {
        if (prices.empty()) {
            return 0;
        }

        int n = prices.size();
        int f0 = -prices[0];
        int f1 = 0;
        int f2 = 0;
        for (int i = 1; i < n; ++i) {
            int newf0 = max(f0, f2 - prices[i]);
            int newf1 = f0 + prices[i];
            int newf2 = max(f1, f2);
            f0 = newf0;
            f1 = newf1;
            f2 = newf2;
        }

        return max(f1, f2);
    }
};
 

312. 戳氣球

Tag: 動態規劃

class Solution {
public:
    int maxCoins(vector<int>& nums) {
        int n = nums.size();
        vector<vector<int>> rec(n + 2, vector<int>(n + 2));
        vector<int> val(n + 2);
        val[0] = val[n + 1] = 1;
        for (int i = 1; i <= n; i++) {
            val[i] = nums[i - 1];
        }
        for (int i = n - 1; i >= 0; i--) {
            for (int j = i + 2; j <= n + 1; j++) {
                for (int k = i + 1; k < j; k++) {
                    int sum = val[i] * val[k] * val[j];
                    sum += rec[i][k] + rec[k][j];
                    rec[i][j] = max(rec[i][j], sum);
                }
            }
        }
        return rec[0][n + 1];
    }
};
 

313. 超級醜數

Tag: 動態規劃

class Solution {
public:
    int nthSuperUglyNumber(int n, vector<int>& primes) {
        vector<long> dp(n + 1);
        int m = primes.size();
        vector<int> pointers(m, 0);
        vector<long> nums(m, 1);
        for (int i = 1; i <= n; i++) {
            long minNum = INT32_MAX;
            for (int j = 0; j < m; j++) {
                minNum = min(minNum, nums[j]);
            }
            dp[i] = minNum;
            for (int j = 0; j < m; j++) {
                if (nums[j] == minNum) {
                    pointers[j]++;
                    nums[j] = dp[pointers[j]] * primes[j];
                }
            }
        }
        return dp[n];
    }
};

316. 去除重複字母

Tag: 單調棧

class Solution {
public:
    string removeDuplicateLetters(string s) {
        vector<int> vis(26), num(26);
        for (char ch : s) {
            num[ch - 'a']++;
        }

        string stk;
        for (char ch : s) {
            if (!vis[ch - 'a']) {
                while (!stk.empty() && stk.back() > ch) {
                    if (num[stk.back() - 'a'] > 0) {
                        vis[stk.back() - 'a'] = 0;
                        stk.pop_back();
                    } else {
                        break;
                    }
                }
                vis[ch - 'a'] = 1;
                stk.push_back(ch);
            }
            num[ch - 'a'] -= 1;
        }
        return stk;
    }
};

318. 最大單詞長度乘積

Tag: 字串 位運算

class Solution {
public:
    bool helper(const string& word1, const string& word2) {
        int n1 = 0, n2 = 0;
        auto f = [](char c) { return c - 'a'; };
        for (auto& c : word1) {
            n1 |= (1 << f(c));
        }
        for (auto& c : word2) {
            n2 |= (1 << f(c));
        }
        return (n1 & n2) == 0;
    }

    int maxProduct(vector<string>& words) {
        int res = 0;
        for (int i = 0; i < words.size(); i++) {
            for (int j = i + 1; j < words.size(); j++) {
                if (helper(words[i], words[j])) {
                    int v = words[i].size() * words[j].size();
                    res = max(res, v);
                }
            }
        }
        return res;
    }
};

319. 燈泡開關

Tag: 數學

class Solution {
public:
    int bulbSwitch(int n) {
        return sqrt(n); //return sqrt(n + 0.5);
    }
};

322. 零錢兌換

Tag: 動態規劃

class Solution {
public:
    int coinChange(vector<int>& coins, int amount) {
        int n = coins.size();
        const int IMPOSSIBLE = amount + 1;
        vector<int> dp(amount + 1, IMPOSSIBLE);
        dp[0] = 0;
        for (int i = 1; i <= amount; i++) {
            for (int j = 0; j < n; j++) {
                if (i >= coins[j]) {
                    dp[i] = min(dp[i], dp[i - coins[j]] + 1);
                }
            }
        }
        return dp[amount] > amount ? -1 : dp[amount];
    }
};

324. 擺動排序 II

Tag: 排序

class Solution {
public:
    void wiggleSort(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        vector<int> sortNums(nums);
        int left = (n - 1) / 2, right = n - 1;
        for(int i = 0; i < n; i++) {
            if((i & 1) == 0) {
                nums[i] = sortNums[left--];
            }else{
                nums[i] = sortNums[right--];
            }
        }

    }
};