leetcode随笔VI

1. leetcode题目
   1.1题目描述
   1.2知识点及思路
   1.3代码
2. 总结

一.leetcode题目
1.Factorial Trailing Zeroes
题目描述：给定一个数求其阶乘当中0的个数
知识点：观察找数学规律；二进制移位操作
思路：①找5的倍数②统计出现次数count
代码如下：

class Solution {public:    int trailingZeroes(int n)     {        int ret=0;        while(n)        {            ret+=n/5;            n=n/5;        }        return ret;    }};

2.Minimum Depth of Binary Tree
题目描述：求解二叉树的最小深度
知识点：树的基本概念；层次遍历；递归求解
思路：①层次遍历二叉树②遇到左右子树全为空，!temp.first->left&&!temp.first->right，则为最小深度 注：用make_pair做层次结构
参考链接：

http://blog.csdn.net/zziymt/article/details/50708255#comments
http://blog.csdn.net/sbitswc/article/details/26526031

代码如下：

/** * 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:    int minDepth(TreeNode* root)     {      if(!root)       return 0;      queue<pair<TreeNode*,int>>vec;      vec.push(make_pair(root,1));      while(!vec.empty())      {          pair<TreeNode*,int>temp=vec.front();          vec.pop();          if(!temp.first->left&&!temp.first->right)            return temp.second;          if(temp.first->left)            vec.push(make_pair(temp.first->left,temp.second+1));          if(temp.first->right)            vec.push(make_pair(temp.first->right,temp.second+1));      }    }};

python代码：

# Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneclass Solution(object):    def minDepth(self, root):        """        :type root: TreeNode        :rtype: int        """        if not root:            return 0        stack,res=[(root,1)],[]        while stack:            p,nDepth=stack.pop()            if not p.left and not p.right:                res.append(nDepth)            if p.right:                stack.append((p.right,nDepth+1))            if p.left:                stack.append((p.left,nDepth+1))        return min(res)

3.Valid Sudoku
题目描述：判断是否为数独
知识点：分治算法
思路：①遍历每行②遍历每列③遍历每个九宫格
知识扩展：回溯法（后面集中讲解）
参考代码如下：

class Solution {public://链接：hackersun007的修行之路    bool isValidSudoku(vector<vector<char>>& board)     {        // Note: The Solution object is instantiated only once.        vector<vector<bool>> rows(9, vector<bool>(9,false));        vector<vector<bool>> cols(9, vector<bool>(9,false));        vector<vector<bool>> blocks(9, vector<bool>(9,false));        for(int i = 0; i < 9; i++)            for(int j = 0; j < 9; j++)            {                if(board[i][j] == '.')continue;                int num = board[i][j] - '1';                if(rows[i][num] || cols[j][num] || blocks[i - i%3 + j/3][num])                    return false;                rows[i][num] = cols[j][num] = blocks[i - i%3 + j/3][num] = true;            }        return true;    }};

二.总结
I.方法虽同，亦需学习他人优点II.让我们一同努力，明天会更好！