leetcode-第十一周

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239. Sliding Window Maximum

class Solution {public:    vector<int> maxSlidingWindow(vector<int>& nums, int k) {        deque<int> q;        vector<int> ret;        for (int i = 0; i < nums.size(); i++) {            while (!q.empty() && nums[i] >= nums[q.back()]) q.pop_back();            q.push_back(i);            while (i >= k && i - k >= q.front()) q.pop_front();            if (i >= k - 1) ret.push_back(nums[q.front()]);        }        return ret;    }};

240. Search a 2D Matrix II

class Solution {private:    int nrow, ncol;public:    bool searchMatrix(vector<vector<int>>& matrix, int target) {        nrow = matrix.size();        if (nrow == 0) return false;        ncol = matrix[0].size();        if (ncol == 0) return false;        for (int r = 0; r < nrow; r++) {            vector<int> &tmp = matrix[r];            int ll = 0, rr = ncol - 1;            while (ll <= rr) {                int mid = ll + (rr - ll) / 2;                if (tmp[mid] > target) rr = mid - 1;                else if (tmp[mid] == target) return true;                else ll = mid + 1;            }        }        return false;    }};

241. Different Ways to Add Parentheses

/** * 思路：tokenize+DFS暴力运算符先后顺序(permutation of operators) */class Solution {private:    vector<string> tokenize(string &input) {        stringstream ss(input);        vector<string> ret;        int tmp; ss >> tmp;        ret.push_back(to_string(tmp));        while (!ss.eof()) {            char c;            ss >> c >> tmp;            string str; str.push_back(c);            ret.push_back(str);            ret.push_back(to_string(tmp));        }        return ret;    }    int calculate(int lhs, string &op, int rhs) {        if (op == "+") return lhs + rhs;        else if (op == "-") return lhs - rhs;        else if (op == "*") return lhs * rhs;    }    // [left, right]    vector<int> dfs(vector<string> &tokens, int left, int right) {        if (left + 1 == right || left > right) return vector<int>(); // 非法输入        if (left == right) return vector<int>{stoi(tokens[left])}; // 终止条件        vector<int> ret;        for (int i = left + 1; i < right; i += 2) {            vector<int> ll = dfs(tokens, left, i - 1);            vector<int> rr = dfs(tokens, i + 1, right);            for (auto l: ll) {                for (auto r: rr) {                    int val = calculate(l, tokens[i], r);                    ret.push_back(val);                }            }        }        return ret;    }public:    vector<int> diffWaysToCompute(string input) {        vector<string> tokens = tokenize(input);        vector<int> ret = dfs(tokens, 0, tokens.size() - 1);        //for (auto t: tokens) cout << t << endl;        sort(ret.begin(), ret.end());        //ret.erase(unique(ret.begin(), ret.end()), ret.end());        return ret;    }};

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