面试笔试杂项积累-leetcode 171-175

171.171-Excel Sheet Column Number-Difficulty: Easy

Related to question Excel Sheet Column Title

Given a column title as appear in an Excel sheet, return its corresponding column number.

For example:

    A -> 1    B -> 2    C -> 3    ...    Z -> 26    AA -> 27    AB -> 28 

思路

与168题恰好相反

public class Solution {    public int TitleToNumber(string s) {                int anwser = 0;        for (int i = 0; i < s.Length; i++)        {            anwser += (s[i] - 64) * (int)Math.Pow(26, s.Length - i - 1);        }        return anwser;    }}

172.172-Factorial Trailing Zeroes-Difficulty: Easy

Given an integer n, return the number of trailing zeroes in n!.

Note: Your solution should be in logarithmic time complexity.

思路

对n!做质因数分解n!=2^x*3^y*5^z*...

显然0的个数等于min(x,z)，并且min(x,z)==z

证明：

对于阶乘而言，也就是1*2*3*...*n
[n/k]代表1~n中能被k整除的个数
那么很显然
[n/2] > [n/5] (左边是逢2增1，右边是逢5增1)
[n/2^2] > [n/5^2](左边是逢4增1，右边是逢25增1)
……
[n/2^p] > [n/5^p](左边是逢2^p增1，右边是逢5^p增1)
随着幂次p的上升，出现2^p的概率会远大于出现5^p的概率。
因此左边的加和一定大于右边的加和，也就是n!质因数分解中，2的次幂一定大于5的次幂

参考:

http://www.cnblogs.com/ganganloveu/p/4193373.html   

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

173.173-Binary Search Tree Iterator-Difficulty: Medium

mplement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.

Calling next() will return the next smallest number in the BST.

Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, whereh is the height of the tree.

思路

对二叉搜索树进行design，next()返回下一个最小值

使用栈来解决

参考：

https://leetcode.com/discuss/78858/my-solution-with-less-than-10-lines-of-code

/** * Definition for binary tree * public class TreeNode { *     public int val; *     public TreeNode left; *     public TreeNode right; *     public TreeNode(int x) { val = x; } * } */public class BSTIterator {    Stack<TreeNode> stack = new Stack<TreeNode>();        public BSTIterator(TreeNode root)        {            pushAll(root);        }        /** @return whether we have a next smallest number */        public bool HasNext()        {            return stack.Count > 0 ? true : false;        }        /** @return the next smallest number */        public int Next()        {            TreeNode temp = stack.Pop();            pushAll(temp.right);            return temp.val;        }        public void pushAll(TreeNode root)        {            while (root != null)            {                stack.Push(root);                root = root.left;            }        }}/** * Your BSTIterator will be called like this: * BSTIterator i = new BSTIterator(root); * while (i.HasNext()) v[f()] = i.Next(); */