# 最长连续回文串Swift实现

## 题目：Longest Palindromic Substring

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example:

``````Input: "babad"

Output: "bab"

Note: "aba" is also a valid answer.
``````

Example:

``````Input: "cbbd"

Output: "bb"
``````

## 算法分析

### 解法二：

• 动态规划，时间复杂度:O(n^2)。
dp[i][j] 表示从i~j的子串是否是回文串。

``````dp[i][j] = {
dp[i-1][j + 1] : s[i] == s[j],
false          : s[i] != s[j]
}
``````

Swift实现

``````class Solution {
func longestPalindrome(_ s: String) -> String {
var dp:[[Bool]] = [];
if s.count <= 1{
return s;
}

var longest:Int = 1;
var left:Int = 0;
var right:Int = 0;

for var i in 0...s.count - 1{
var eachRow:[Bool] = [];
for var j in 0...s.count - 1{
if i == j{
eachRow.append(true);
}else{
eachRow.append(false);
}
}
dp.append(eachRow);
}

var i:Int = 0;
var j:Int = 0;
for var character_j in s {
if j == 0 {
j += 1;
continue;
}
i = 0;
for var character_i in s {
if character_i == character_j {
dp[i][j] = dp[i + 1][j - 1] || j - i <= 1;
if dp[i][j] && j - i + 1 > longest{
longest = j - i + 1;
left = i;
right = j;
}
}else{
dp[i][j] = false;
}
i += 1;
if i >= j{
break;
}
}
j += 1;
}
let leftIndex = s.index(s.startIndex, offsetBy: left);
let rightIndex = s.index(s.startIndex, offsetBy: right);
return String(s[leftIndex...rightIndex]);
}
}
``````

AC结果：可能是swift处理字符串的效率问题会超时。

``````Time Limit Exceeded
``````

### 第三种解法

• Manacher算法,时间复杂度：O(n)。

Manacher's ALGORITHM: O(n)时间求字符串的最长回文子串思路分析这里介绍的很清晰，

#### 首先解决回文长度为奇偶的问题

• 插桩处理，整个字符串的前后间隔处插入'#'字符，最终得到的字符串就一定是奇数长度，回文的长度也一定是奇数长度。

``````# c # b # b # d #
1 2 1 2 3 2 1 2 1
``````

#### 解决计算`P[i]`问题

Longest Palindromic Substring_1.png

Longest Palindromic Substring_2.png

``````//记j = 2 * id - i，也就是说 j 是 i 关于 id 的对称点(j = id - (i - id))
P[i] = {
P[j] ,mx - i > P[j]
mx - i, mx - i <= P[j]
}
``````

#### Swift 实现

``````class Solution {
func longestPalindrome(_ s: String) -> String {
if s.count <= 1 {
return s;
}

// 1.间隔之间先插入#
var newString:String = "#";
for var character in s {
newString.append(character);
newString = newString + "#";
}
let characters = Array(newString);

// 2.遍历找出以每个节点作为轴最长半径
var maxId:Int = 0;
var max:Int = 0;
var ids:[Int] = [];
ids.append(1);
var maxLength:Int = 1;
var maxLengthIndex = 0;

for var i in 1...characters.count - 1 {
var j:Int = maxId - (i - maxId);
if max > i && j >= 0 {
ids.append(min(ids[j], max - i));
}else{
ids.append(1);
}
while i + ids[i] <= characters.count - 1 && i - ids[i] >= 0 && characters[i + ids[i]] == characters[i - ids[i]]{
ids[i] += 1;
}

if i + ids[i] - 1 > max {
maxId = i;
max = i + ids[i] - 1;
}

if ids[i] > maxLength{
maxLength = ids[i];
maxLengthIndex = i;
}
}
let leftIndex = s.index(s.startIndex, offsetBy: (maxLengthIndex - (maxLength - 1))/2);
let rightIndex = s.index(leftIndex, offsetBy:maxLength - 1 - 1);
return String(s[leftIndex...rightIndex]);
}
}
``````

AC结果：

``````Accepted
``````
The text passage describes a programming problem and its solutions in Swift, specifically for finding the longest palindromic substring in a given string. It includes examples, algorithm analysis, and Swift code implementations.
no_question

the passage does not contain a mathematical question or provide an answer to one. it is focused on a programming problem and its solutions.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# 最长连续回文串Swift实现

## 题目：Longest Palindromic Substring

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example:

``````Input: "babad"

Output: "bab"

Note: "aba" is also a valid answer.
``````

Example:

``````Input: "cbbd"

Output: "bb"
``````

## 算法分析

### 解法二：

• 动态规划，时间复杂度:O(n^2)。
dp[i][j] 表示从i~j的子串是否是回文串。

``````dp[i][j] = {
dp[i-1][j + 1] : s[i] == s[j],
false          : s[i] != s[j]
}
``````

Swift实现

``````class Solution {
func longestPalindrome(_ s: String) -> String {
var dp:[[Bool]] = [];
if s.count <= 1{
return s;
}

var longest:Int = 1;
var left:Int = 0;
var right:Int = 0;

for var i in 0...s.count - 1{
var eachRow:[Bool] = [];
for var j in 0...s.count - 1{
if i == j{
eachRow.append(true);
}else{
eachRow.append(false);
}
}
dp.append(eachRow);
}

var i:Int = 0;
var j:Int = 0;
for var character_j in s {
if j == 0 {
j += 1;
continue;
}
i = 0;
for var character_i in s {
if character_i == character_j {
dp[i][j] = dp[i + 1][j - 1] || j - i <= 1;
if dp[i][j] && j - i + 1 > longest{
longest = j - i + 1;
left = i;
right = j;
}
}else{
dp[i][j] = false;
}
i += 1;
if i >= j{
break;
}
}
j += 1;
}
let leftIndex = s.index(s.startIndex, offsetBy: left);
let rightIndex = s.index(s.startIndex, offsetBy: right);
return String(s[leftIndex...rightIndex]);
}
}
``````

AC结果：可能是swift处理字符串的效率问题会超时。

``````Time Limit Exceeded
``````

### 第三种解法

• Manacher算法,时间复杂度：O(n)。

Manacher's ALGORITHM: O(n)时间求字符串的最长回文子串思路分析这里介绍的很清晰，

#### 首先解决回文长度为奇偶的问题

• 插桩处理，整个字符串的前后间隔处插入'#'字符，最终得到的字符串就一定是奇数长度，回文的长度也一定是奇数长度。

``````# c # b # b # d #
1 2 1 2 3 2 1 2 1
``````

#### 解决计算`P[i]`问题

Longest Palindromic Substring_1.png

Longest Palindromic Substring_2.png

``````//记j = 2 * id - i，也就是说 j 是 i 关于 id 的对称点(j = id - (i - id))
P[i] = {
P[j] ,mx - i > P[j]
mx - i, mx - i <= P[j]
}
``````

#### Swift 实现

``````class Solution {
func longestPalindrome(_ s: String) -> String {
if s.count <= 1 {
return s;
}

// 1.间隔之间先插入#
var newString:String = "#";
for var character in s {
newString.append(character);
newString = newString + "#";
}
let characters = Array(newString);

// 2.遍历找出以每个节点作为轴最长半径
var maxId:Int = 0;
var max:Int = 0;
var ids:[Int] = [];
ids.append(1);
var maxLength:Int = 1;
var maxLengthIndex = 0;

for var i in 1...characters.count - 1 {
var j:Int = maxId - (i - maxId);
if max > i && j >= 0 {
ids.append(min(ids[j], max - i));
}else{
ids.append(1);
}
while i + ids[i] <= characters.count - 1 && i - ids[i] >= 0 && characters[i + ids[i]] == characters[i - ids[i]]{
ids[i] += 1;
}

if i + ids[i] - 1 > max {
maxId = i;
max = i + ids[i] - 1;
}

if ids[i] > maxLength{
maxLength = ids[i];
maxLengthIndex = i;
}
}
let leftIndex = s.index(s.startIndex, offsetBy: (maxLengthIndex - (maxLength - 1))/2);
let rightIndex = s.index(leftIndex, offsetBy:maxLength - 1 - 1);
return String(s[leftIndex...rightIndex]);
}
}
``````

AC结果：

``````Accepted
``````
user
The text passage describes a programming problem and its solutions in Swift, specifically for finding the longest palindromic substring in a given string. It includes examples, algorithm analysis, and Swift code implementations.

TYPE: NO_QUESTION

The passage does not contain a mathematical question or provide an answer to one. It is focused on a programming problem and its solutions.