Leecode05. Longest palindrome substring -- leecode100 question series

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Problem description

Give you a string s, find s The longest palindrome substring in .

Example 1：
Input ：s = “babad”
Output ：“bab”
explain ：“aba” It's the same answer .

Example 2：
Input ：s = “cbbd”
Output ：“bb”

Example 3：

Input ：s = “a”
Output ：“a”

Example 4：
Input ：s = “ac”
Output ：“a”

Tips ：
1 <= s.length <= 1000
s Just numbers and letters （ Capital and / Or lowercase ） form

Ideas

Center detection method （ Common methods for solving palindrome strings ）

Complexity analysis

Time complexity ：$O(n^2)$, among $n$ Is the length of the string . The length is $1$ and $2$ The palindrome centers are $n$ and $n-1$ individual , Each palindrome center will expand outward at most $O(n)$ Time .

Spatial complexity ：$O(1)$.

Code 1 （ Universal code ）

class Solution {

public:
string longestPalindrome(string s) {

int left = 0, right = s.length() - 1;
int maxLen = 0, nowLen = 0;
string res = "";
for(int i = 0; i < s.length(); i++) {

// Palindrome sequence is odd 
nowLen = 1;
int l = i-1, r = i+1;
while (l >= left && r <= right) {

if(s[l] == s[r]) {

nowLen += 2;
} else {

break;
}
l--; r++;
}
if(nowLen > maxLen) {

maxLen = nowLen;
res = s.substr(l+1, maxLen);
}
// Palindrome sequence is even 
nowLen = 0;
l = i, r = i+1;
while (l >= left && r <= right) {

if(s[l] == s[r]) {

nowLen += 2;
} else {

break;
}
l--; r++;
}
if(nowLen > maxLen) {

maxLen = nowLen;
res = s.substr(l+1, maxLen);
}
}
return res;
}
};

Code 2： Function reuse

class Solution {

public:
int func(int& nowLen, int l, int r, string s) {

while (l >= 0 && r <= s.length()) {

if(s[l] == s[r]) {

nowLen += 2;
} else {

break;
}
l--; r++;
}
return l;
}
string longestPalindrome(string s) {

int maxLen = 0, nowLen, sta;
string res = "";
for(int i = 0; i < s.length(); i++) {

// Palindrome sequence is odd 
nowLen = 1;
sta = func(nowLen, i-1, i+1, s);
if(nowLen > maxLen) {

maxLen = nowLen;
res = s.substr(sta+1, maxLen);
}
// Palindrome sequence is even 
nowLen = 0;
sta = func(nowLen, i, i+1, s);
if(nowLen > maxLen) {

maxLen = nowLen;
res = s.substr(sta+1, maxLen);
}
}
return res;
}
};

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