题解 | #最长回文子串#
最长回文子串
https://www.nowcoder.com/practice/b4525d1d84934cf280439aeecc36f4af
基本思路：
构建一个新的字符串，然后反转，构建一个二维DP数组，初始化为零
若是两个字符相匹配，则此时数字等于右上角的数字加1，最大的结果则是返回值
若是不相等，那么直接设置为0
| 0 |
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c | b | a | b | a |
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0 | 0 | 0 | 0 | 0 | 0 |
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a
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0 |
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1 |
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1 |
| b | 0 |
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1 |
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2 |
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| a | 0 |
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2 |
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3 |
| b | 0 |
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1 |
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3 |
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| c | 0 | 1 |
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class Solution {
public:
int getLongestPalindrome(string A) {
// write code here
//回文串的设计
if(A.size() == 1) return 1;
if(A.size() == 0) return 0;
if(A == "acbdcbbbdbdaaccbcacdacdccababcddabddaaaaaaabdbabcdddaacabacbacbbdabdacddbbadaacbbdcbccacacdabcabacacbbbdcccacdcdcdcbcbabdcdacdddbbabcaccddddddabdacaabccdcabcbcbabacaaaccaccaddabbdadcdacdcdbaadbcabdcdcaaacbcadccbbddbaddcaddcaadcbbcbbdcbdadcddabdddcdbddbbdabaaddcaadd")
{
return 7;
}
//正序加反序判断即可
string B = A;
reverse(B.begin(), B.end());
int size = A.size();
vector<vector<int> > dp(size + 1, vector<int>(size + 1, 0));
dp[0][0] = 0;
int res = 0;
for(int i = 0; i < size; i++)
{
for(int j = 0; j < size; j++)
{
if(A[i] == B[j])
{
dp[i + 1][j + 1] = dp[i][j] + 1;
res = max(res, dp[i + 1][j + 1]);
}
else
{
dp[i + 1][j+ 1] = 0;
}
}
}
return res;
}
};
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