# README
"love",
"hate",
"yes"
}
输出: ["like", "love", "hate"]
### **Challenge**
遍历两次的办法很容易想到，如果只遍历一次你有没有什么好办法？
[传送门](./133-longestword.go)
## 135-combinationsum
### **Description**
中文English
给定一个候选数字的集合 `candidates` 和一个目标值 `target`. 找到 `candidates` 中所有的和为 `target` 的组合.
在同一个组合中, `candidates` 中的某个数字不限次数地出现.
所有数值 (包括 `target` ) 都是正整数.返回的每一个组合内的数字必须是非降序的.返回的所有组合之间可以是任意顺序.解集不能包含重复的组合.
### **Example**
**样例 1:**
输入: candidates = [2, 3, 6, 7], target = 7 输出: [[7], [2, 2, 3]]
**样例 2:**
输入: candidates = [1], target = 3 输出: [[1, 1, 1]]
[传送门](./135-combinationsum.go)
## 1324-countprimes
```cgo
计算小于非负数n的质数的个数。
Example
样例 1
输入: n = 2
输出: 0
样例 2
输入: n = 4
输出: 2
解析：2, 3 是素数
# Functions
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给你一个整数n.
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左子树 ---> 根结点 ---> 右子树BFS.
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O(nlogn) 稳定.
O(n+mlogm) 超时,但是不稳定,最差的是O(n^2).
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dp[i][j]表示A串匹配到i，B串匹配到j时的最大公共长度
dp[i][j]=dp[i-1][j-1]+1 ,A[i]==B[j]
dp[i][j]=0 ,A[i]!=B[j]
*/.
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a,b必须已经是排序的的.
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超时.
原理:
$$
f(n)=[\farc{n!}{5}] + [\farc{n!}{5^2}] +\cdots + [\farc{n!}{5^n}]
$$
$$
q_0 = n
q_{i+1} = [\farc{q_i}{5}]
$$
$$
q_{i+1} = 0 <==> 5^{k+1} > n
$$
for q != 0 {
q /= 5 // ==> a1 + a2 + ..
O(N).
O(N^2).
设计一个算法，找出只含素因子2，3，5 的第 n 小的数。
符合条件的数如：1, 2, 3, 4, 5, 6, 8, 9, 10, 12..