LeetCode40-49題解

43.Multiply Strings
思路：就是按照我們手算乘法的步驟來解決的，可以看下圖，要注意一點下標的標記。

string multiply(string num1, string num2) {
        int n1 = num1.size();
        int n2 = num2.size();
        vector<int> temp(n1+n2,0);
        int start = n1 + n2 - 1;
        int carry = 0;
        string str;
        for (int i = 0; i < n1; i++) {
            for (int j = 0; j < n2; j++) {
                temp[i + j] += (num1[n1 - i - 1] - '0' ) * (num2[n2 - j - 1] - '0');
            }
        }
        for (int i = 0; i < n1 + n2; i++) {
            temp[i] = temp[i] + carry;
            carry = temp[i] / 10;
            temp[i] = temp[i] % 10;
        }
        while (!temp[start])
            start--;
        if (start < 0)
            return "0";
        while (start >= 0)
            str += temp[start--] + '0';
        return str;


    }

46.Permutations
思路：很明顯，要用DFS。

vector<vector<int>> permute(vector<int>& nums) {
        vector<int> visited(nums.size(),0);
        vector<vector<int>> res;
        vector<int> out;
        DFS(nums,0,visited,out,res);
        return res;
    }
    void DFS(vector<int>& nums,int level, vector<int>& visited,vector<int>& out,vector<vector<int>>& res) {
        if (level == nums.size())
            res.push_back(out);
        else {
            for (int i = 0; i < nums.size(); i++) {
                if (visited[i] == 0) {
                    visited[i] = 1;
                    out.push_back(nums[i]);
                    DFS(nums,level + 1,visited,out,res);
                    out.pop_back();
                    visited[i] = 0;
                }

            }
        }
    }

47.Permutations II
思路：和46題一樣，當時需要判斷重複數字項。判斷依據是重複數字只有前面的數用過以後，也就是說visited = 1之後，後面的數才能用。

 vector<vector<int>> permuteUnique(vector<int>& nums) {
        sort(nums.begin(),nums.end());
        vector<vector<int>> res;
        vector<int> out;
        vector<int> visited(nums.size(),0);
        DFS(nums,0,visited,out,res);
        return res;
    }
    void DFS(vector<int>& nums,int level,vector<int>& visited,vector<int>& out,vector<vector<int>>& res) {
        if (level == nums.size())
            res.push_back(out);
        else {
            for (int i = 0; i < nums.size(); i++) {
                if (i > 0 && nums[i] == nums[i - 1] && visited[i - 1] == 0)
                    continue;
                if (visited[i] == 0) {
                    visited[i] = 1;
                    out.push_back(nums[i]);
                    DFS(nums,level+1,visited,out,res);
                    out.pop_back();
                    visited[i] = 0;
                }
            }
        }
    }