题目描述
The count-and-say sequence is the sequence of integers with the first five terms as following:
1. 1
2. 11
3. 21
4. 1211
5. 111221
1 is read off as "one 1" or 11.
11 is read off as "two 1s" or 21.
21 is read off as "one 2, then one 1" or 1211.
Given an integer n where 1 ≤ n ≤ 30, generate the nth term of the count-and-say sequence. You can do so recursively, in other words from the previous member read off the digits, counting the number of digits in groups of the same digit.
Note: Each term of the sequence of integers will be represented as a string.
Example 1: Input: 1 Output: "1"
Explanation: This is the base case.
Example 2:
Input: 4 Output: "1211"
Explanation: For n = 3 the term was "21" in which we have two groups "2" and "1", "2" can be read as "12" which means frequency = 1 and value = 2, the same way "1" is read as "11", so the answer is the concatenation of "12" and "11" which is "1211".
解题思路1: 暴力破解
这个题目主要是理解题意, 下一次的结果是根据上一次的结果来进行读数. 所以我们只需要每次根据上一次的结果进行遍历, 得到下一次的结果
时间复杂度: O(n^2)
示例代码1
func countAndSay(_ n: Int) -> String {
var result = "1"
for _ in 1..<n {
var count = 0
let strArr = Array(result)
var tempStr = ""
var tempC = strArr.first!
for (_, c) in strArr.enumerated() {
if c == tempC {
count = count + 1
}else {
tempStr = tempStr + "\(count)\(tempC)"
tempC = c
count = 1
}
}
result = tempStr + "\(count)\(tempC)"
}
return result
}
解题思路2: 递归法
方案2: 既然当前结果跟上次的结果有关系, 我们很容易就想到用递归来解决
时间复杂度: O(n^2)
示例代码2:
func countAndSay(_ n: Int) -> String {
return getCurrentStr(allNum: n, current: 1, lastStr: "1")
}
func getCurrentStr(allNum: Int, current: Int, lastStr: String) -> String{
if current < allNum {
var count = 0
let strArr = Array(lastStr)
var tempStr = ""
var tempC = strArr.first!
for (_, c) in strArr.enumerated() {
if c == tempC {
count = count + 1
}else {
tempStr = tempStr + "\(count)\(tempC)"
tempC = c
count = 1
}
}
tempStr = tempStr + "\(count)\(tempC)"
return getCurrentStr(allNum: allNum, current: current+1, lastStr: tempStr)
}else {
return lastStr
}
}
