[Swift]LeetCode1312. 让字符串成为回文串的最少插入次数 | Minimum Insertion Steps to Make a String Palindrome

★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★
➤微信公众号：山青咏芝（let_us_code）
➤博主域名：https://www.zengqiang.org
➤GitHub地址：https://github.com/strengthen/LeetCode
➤原文地址：https://www.cnblogs.com/strengthen/p/12151983.html
➤如果链接不是山青咏芝的博客园地址，则可能是爬取作者的文章。
➤原文已修改更新！强烈建议点击原文地址阅读！支持作者！支持原创！
★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★★

Given a string s. In one step you can insert any character at any index of the string.

Return the minimum number of steps to make s palindrome.

A Palindrome String is one that reads the same backward as well as forward.

 

Example 1:

Input: s = "zzazz"
Output: 0
Explanation: The string "zzazz" is already palindrome we don't need any insertions.

Example 2:

Input: s = "mbadm"
Output: 2
Explanation: String can be "mbdadbm" or "mdbabdm".

Example 3:

Input: s = "leetcode"
Output: 5
Explanation: Inserting 5 characters the string becomes "leetcodocteel".

Example 4:

Input: s = "g"
Output: 0

Example 5:

Input: s = "no"
Output: 1

 

Constraints:

• 1 <= s.length <= 500
• All characters of s are lower case English letters.

---

给你一个字符串 s ，每一次操作你都可以在字符串的任意位置插入任意字符。

请你返回让 s 成为回文串的 最少操作次数 。

「回文串」是正读和反读都相同的字符串。

 

示例 1：

输入：s = "zzazz"
输出：0
解释：字符串 "zzazz" 已经是回文串了，所以不需要做任何插入操作。

示例 2：

输入：s = "mbadm"
输出：2
解释：字符串可变为 "mbdadbm" 或者 "mdbabdm" 。

示例 3：

输入：s = "leetcode"
输出：5
解释：插入 5 个字符后字符串变为 "leetcodocteel" 。

示例 4：

输入：s = "g"
输出：0

示例 5：

输入：s = "no"
输出：1

 

提示：

• 1 <= s.length <= 500
• s 中所有字符都是小写字母。

---

Runtime: 300 ms

Memory Usage: 21.9 MB

 1 class Solution {
 2     var memo:[[Int]]!
 3     func minInsertions(_ s: String) -> Int {
 4         let s = Array(s)
 5         memo = [[Int]](repeating: [Int](repeating: -1, count: s.count), count: s.count)
 6         return dp(s,0,s.count - 1)
 7     }
 8     
 9     func dp(_ s:[Character],_ i:Int,_ j:Int) -> Int
10     {
11         //Base case.
12         if i >= j
13         {
14             return 0
15         }
16         //Check if we have already calculated the value for the pair `i` and `j`.
17         if memo[i][j] != -1
18         {
19             return memo[i][j]
20         }
21         //Recursion as mentioned above.
22         memo[i][j] = s[i] == s[j] ? dp(s,i+1,j-1) : 1 + min(dp(s,i+1,j),dp(s,i,j-1))
23         return memo[i][j]
24     }
25 }