leetcode weekly contest 43

leetcode649. Dota2 Senate

leetcode649.Dota2 Senate

思路：

模拟规则round by round。

class Solution {
public:
    string predictPartyVictory(string senate) {
        int len = senate.size();
        int r = 0;
        int d = 0;
        int i = 0;
        while(1)
        {
            if(senate[i] == 'R')
            {
                int j = 1;
                while(senate[(i + j) % len] != 'D' && j < len) j++;
                if(senate[(i + j) % len] == 'R') return "Radiant";
                else senate[(i + j) % len] = '#';
            }
            else if(senate[i] == 'D')
            {
                int j = 1;
                while(senate[(i + j) % len] != 'R' && j < len) j++;
                if(senate[(i + j) % len] == 'D') return "Dire";
                else senate[(i + j) % len] = '#';
            }
            i = (i + 1) % len;
        }
        
    }
};

leetcode650. 2 Keys Keyboard

思路：

找到最大的约数，dp。
注意copy算一步。

class Solution {
public:
    int find(int x)
    {
        for(int i = 2; i <= sqrt(x); i++)
        {
            if(x % i == 0) return i;
        }
        return x;
    }
    int minSteps(int n) {
        vector<int> res(n + 1, 0);
        int temp;
        for(int i = 2; i <= n; i++)
        {
            temp = find(i);
            res[i] = res[i / temp] + temp;
        }
        return res[n];
    }
};

leetcode651. 4 Keys Keyboard

思路：

dp.
res[n] = max{res[n - 1] + 1, res[n - 5] * 4, res[n - 4] * 3, res[n - 3] * 2};
时间O(n),空间O(1)

class Solution {
public:
    int maxA(int N) {
        //vector<int> res(N + 1);
        int four,three,two,one,five,res;
        five = four = three = two = one = res = 0;
        for(int i = 1; i <= N; i++)
        {
            res = max(one + 1,max(four * 3, max(three * 2,five * 4)));
            five = four;
            four = three;
            three = two;
            two = one;
            one = res;            
        }
        return res;
    }
};

leetcode652. Find Duplicate Subtrees

思路：

这个思路我感觉特别地棒。既然是找duplicates，可以直接用map。吧树给serialize了变成一个string，让map自己比较。然后找到全部的duplicates。
时间O(n)

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    unordered_map<string,vector<TreeNode*> > map;
    vector<TreeNode*> findDuplicateSubtrees(TreeNode* root) {
        vector<TreeNode*> res;
        serialize(root);
        unordered_map<string,vector<TreeNode*>>::iterator it = map.begin();
        while(it != map.end())
        {
            if(it->second.size() > 1 && it->first != "")
            {
                res.push_back(it->second[0]);
            }
            it++;
        }
        return res;
    }
private:
    string serialize(TreeNode* root)
    {
        if(!root) return "";
        string s = '(' + serialize(root->left) + to_string(root->val) + serialize(root->right) + ')';
        map[s].push_back(root);
        return s;
    }
};