【LeetCode】5. Longest Palindromic Substring

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, and there exists one unique longest palindromic substring.

同Palindrome Partitioning II 几乎一致，二维数组动态规划问题。

设置bool数组isPalin

isPalin[i][j]意为s[i...j]是否为回文字符串。

因此很容易找到递推关系：当s[i]==s[j]时

（1）如果i+1==j，或者

（2）如果isPalin[i-1][j+1]

那么isPalin[i][j]=true （此时与最大长度longest进行比较，并记录起始位置i）

否则isPalin[i][j]=false

提高效率需要注意两点：

1、使用数组比使用vector of vector节省时间；

2、每次发现当前最长回文串就使用substr取出会耗时间，改成记录开始位置和长度。

class Solution {
public:
string longestPalindrome(string s) {
if(s == "" || s.size() == 1)
return s;
int size = s.size();
int begin = 0;
int longest = 1;
bool isPalin[1000][1000] = {false};

for(int i = 0; i < size; i ++)
isPalin[i][i] = true;
for(int i = size-2; i >= 0; i --)
{//isPalin[size-1][size-1] has set to true
for(int j = i+1; j < size; j ++)
{//isPalin[i][i] has set to true
//check whether s[i...j] is a palindrome
if(s[i] == s[j])
{
if(j == i+1 || isPalin[i+1][j-1])
{//case1: j is next to i
//case2: s[i+1...j-1] is a palindrome
isPalin[i][j] = true;
if(j-i+1 > longest)
{
longest = j-i+1;
begin = i;
}
}
}
}
}
return s.substr(begin, longest);
}
};