Leetcode -- Longest Palindromic Substring
2015-07-30 00:17
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题目: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.
这个问题,跟LCS的解题思路上非常相似,利用矩阵来记录之前的判断信息,依次来加速后续的判断过程。
大致思路是:
用数组arr[i][j]记录下标从i到j的字串是否为回文。i从后往前走,j从j=i开始,往字符串的尾巴走(下面思路是这个走法)。当然,i从头往后走,j从j=i开始,往字符串的开头走也是可以的。
对于arr[i][j], 若存在s[i] = s[j], 并且j <= i +2 或者s[i+1] = s[j-1], 则arr[i][j]表示的字串是回文。
C++代码:
class Solution {
bool arr[1000][1000];
public:
string longestPalindrome(string s) {
memset(arr, 0, sizeof(arr));
int maxLen = 0;
string str;
for (int i = s.size(); i >= 0; i--)
{
for(int j =i; j < s.size(); j++)
{
if(s[i] == s[j] && ((j -i)<=2 || arr[i+1][j-1]))
{
arr[i][j] = true;
if((j-i+1) > maxLen)
{
maxLen = j -i + 1;
str = s.substr(i, j - i + 1);
}
}
}
}
return str;
}
};
注意:一个char是一个回文串。即‘a’是一个回文哦。
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.
这个问题,跟LCS的解题思路上非常相似,利用矩阵来记录之前的判断信息,依次来加速后续的判断过程。
大致思路是:
用数组arr[i][j]记录下标从i到j的字串是否为回文。i从后往前走,j从j=i开始,往字符串的尾巴走(下面思路是这个走法)。当然,i从头往后走,j从j=i开始,往字符串的开头走也是可以的。
对于arr[i][j], 若存在s[i] = s[j], 并且j <= i +2 或者s[i+1] = s[j-1], 则arr[i][j]表示的字串是回文。
C++代码:
class Solution {
bool arr[1000][1000];
public:
string longestPalindrome(string s) {
memset(arr, 0, sizeof(arr));
int maxLen = 0;
string str;
for (int i = s.size(); i >= 0; i--)
{
for(int j =i; j < s.size(); j++)
{
if(s[i] == s[j] && ((j -i)<=2 || arr[i+1][j-1]))
{
arr[i][j] = true;
if((j-i+1) > maxLen)
{
maxLen = j -i + 1;
str = s.substr(i, j - i + 1);
}
}
}
}
return str;
}
};
注意:一个char是一个回文串。即‘a’是一个回文哦。
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