从零开始的刷LeetCode生活 第17期 171-180

class Solution {
public:
    int titleToNumber(string s) {
        int num = 0;
        for(int i = 0; i < s.size(); i ++)
        {
            num = num * 26 + (s[i] - 'A' + 1);
        }
        return num;
    }
};

class Solution {
public:
    int trailingZeroes(int n) {
        return n < 5 ? 0 : n / 5 + trailingZeroes(n/5);
    }
};

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class BSTIterator {
public:
    stack<TreeNode*> stk;
    BSTIterator(TreeNode* root) {
        while(root)
        {
            stk.push(root);
            root = root->left;
        }
    }
    
    /** @return the next smallest number */
    int next() {
        auto t = stk.top();
        stk.pop();
        for(auto p = t->right; p; p = p->left) stk.push(p);
        return t->val;
    }
    
    /** @return whether we have a next smallest number */
    bool hasNext() {
        return !stk.empty();
    }
};

/**
 * Your BSTIterator object will be instantiated and called as such:
 * BSTIterator* obj = new BSTIterator(root);
 * int param_1 = obj->next();
 * bool param_2 = obj->hasNext();
 */

class Solution {
public:
   int calculateMinimumHP(vector<vector<int>>& dungeon) {
       int n = dungeon.size();
       int m = dungeon[0].size();
       vector<vector<int>> f(n, vector<int>(m, 0));
       for(int i = n - 1; ~i; i --)
           for(int j = m - 1; ~j; j --)
               if(i == n - 1 && j == m - 1)
                   f[i][j] = max(1 - dungeon[i][j], 1);
               else if(i == n - 1)
                   f[i][j] = max(1, f[i][j + 1] - dungeon[i][j]);
               else if(j == m - 1)
                   f[i][j] = max(1, f[i + 1][j] - dungeon[i][j]);
               else 
                   f[i][j] = min(max(1, f[i + 1][j] - dungeon[i][j]), max(1, f[i][j + 1] - dungeon[i][j]));
       return f[0][0];
   }
};

# Write your MySQL query statement below
select FirstName, LastName, City, State
from Person left join Address
on Person.PersonId = Address.PersonId;

class Solution {
public:
    static bool cmp(int x, int y)
    {
        string sx = to_string(x), sy = to_string(y);
        return sx + sy > sy + sx;
    }
    string largestNumber(vector<int>& nums) {
        sort(nums.begin(), nums.end(), cmp);
        string res = "";
        for(auto c : nums)
            res += to_string(c);
        for(int i = 0; i < res.size(); i ++) //去除前面多余的0
            if(i == res.size() - 1 || res[i] != '0')
                return res.substr(i, res.size() - i);
        return res;
    }
};