171. Excel Sheet Column Number
class Solution {
public int titleToNumber(String s) {
if(s==null || s.length()==0)
return 0;
int res = 0;
for(int i = 0;i172. Factorial Trailing Zeroes
阶乘后面0的个数是有递推公式的：
class Solution {
public int trailingZeroes(int n) {
return n < 5 ? 0 : n / 5 + trailingZeroes(n / 5);
}
}
173. Binary Search Tree Iterator
用一个Stack记录从根节点到当前节点的路径。next的时候就返回Stack最上面的元素。不过拿出最上面的元素后，我们还要看一下这个被返回的元素是否有右节点，如果有的话，就把它的右节点及右节点的所有左边节点都压入栈中。另外，初始化栈时，我们要找到最左边的节点，也就是中序遍历的第一个节点，并把根到第一个节点的路径都压入栈。
/**
* Definition for binary tree
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class BSTIterator {
Stack stack;
public BSTIterator(TreeNode root) {
stack = new Stack();
if(root!=null){
TreeNode p = root;
while(p!=null){
stack.push(p);
p = p.left;
}
}
}
/** @return whether we have a next smallest number */
public boolean hasNext() {
return !stack.empty();
}
/** @return the next smallest number */
public int next() {
if(hasNext()){
TreeNode top = stack.pop();
TreeNode p = top.right;
while(p!=null){
stack.push(p);
p = p.left;
}
return top.val;
}
return 0;
}
}
/**
* Your BSTIterator will be called like this:
* BSTIterator i = new BSTIterator(root);
* while (i.hasNext()) v[f()] = i.next();
*/
174. Dungeon Game
这里使用动态规划的思路：
1、如果考虑从左上角到右下角的思路，要考虑路径上有多少药，而且路径上所有的地方的血量都不能小于1，这个需要考虑的因素太多，所以用另一种思路，即从右下角到左上角。
2、要考虑任意位置的最小血量是1.
class Solution {
public int calculateMinimumHP(int[][] dungeon) {
if(dungeon==null || dungeon.length==0)
return 1;
int m = dungeon.length;
int n = dungeon[0].length;
int[][] minHp = new int[m][n];
for(int i = m-1;i>=0;i--){
for(int j=n-1;j>=0;j--){
if(i==m-1 && j==n-1)
minHp[i][j] = 1 + (dungeon[i][j] < 0?Math.abs(dungeon[i][j]):0);
else if(i==m-1)
minHp[i][j] = Math.max(minHp[i][j+1] - dungeon[i][j],1);
else if(j==n-1)
minHp[i][j] = Math.max(1,minHp[i+1][j] - dungeon[i][j]);
else
minHp[i][j] = Math.max(1,Math.min(minHp[i][j+1]-dungeon[i][j],minHp[i+1][j]-dungeon[i][j]));
}
}
return minHp[0][0];
}
}
175. Combine Two Tables
偶尔做一道数据库的题陶冶下情操，这题考查的是左链接：
# Write your MySQL query statement below
select
FirstName,LastName,City,State
from
(select * from Person) t1
left join
(select * from Address) t2
on t1.PersonId = t2.PersonId
176. Second Highest Salary
# Write your MySQL query statement below
select
max(Salary) as SecondHighestSalary
from
Employee
where
Salary < (select Max(Salary) from Employee)
177. Nth Highest Salary
这里考察limit的应用，第一个参数是起始位置，后面是取几条
CREATE FUNCTION getNthHighestSalary(N INT) RETURNS INT
BEGIN
DECLARE M INT;
Set M=N-1;
RETURN (
# Write your MySQL query statement below.
SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT M, 1
);
END
178. Rank Scores
两个表作笛卡尔积，然后过滤掉左表的Score比右表Score小的记录，此时对于左表的每个ID来说，留下的记录都是比自己大或者相等的记录，然后按照左表的ID进行分组，统计比ID对应的Score大的Score有几个。最后不要忘记排序哟。
# Write your MySQL query statement below
select
t1.Score,count(distinct t2.Score) as Rank
from
(select * from Scores) t1
join
(select * from Scores) t2
where t1.Score <= t2.Score
group by t1.Id,t1.Score
order by t1.Score desc
179. Largest Number
这个题是要设计一个排序规则，对于两个字符串来说，如果a+b>b+a，那么a要排在b前面。
这里要介绍一下compareTo函数和Arrays.sort()方法。
compareTo：如果指定的数与参数相等返回0。如果指定的数小于参数返回 -1。如果指定的数大于参数返回 1。
Arrays.sort()：默认是升序，compare方法返回负数时代表obj1 < obj2 ，compare方法返回0时代表obj1 = obj2 ，compare方法返回正数时代表obj1 > obj2
那么也就是说，a排在b前面的规则就是方法,compare方法返回一个负数，那么也就是说b+a 这里还要注意全0的情况，比如数组是两个0，那么返回结果只能有一个0 两次内连接，同时要做一次去重。class Solution {
private class LargerNumberComparator implements Comparator180. Consecutive Numbers
# Write your MySQL query statement below
select
t1.Num as ConsecutiveNums
from
(select * from Logs) t1
inner join
(select * from Logs) t2
on t1.Id = t2.Id - 1
inner join
(select * from Logs) t3
on t1.Id = t3.Id - 2
where
t1.Num = t2.Num and t1.Num = t3.Num
group by t1.Num