1. 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.
2. 思路
查找最长回文子串。
dp算法:
dp[i, j] = dp[i+1, j-1] + 2, if s[i] == s[j];
dp[i, i] = 1
dp[i, i-1] = 0;
迭代求解到最大值,并在过程中记录下起终的下标。
3. 代码
耗时:203ms
class Solution {
public:
void reset(int len) {
for (int i = 0 ; i < len; i++) {
dp[i][i] = true;
for (int j = i + 1; j < len; j++) {
dp[i][j] = false;
}
}
}
string longestPalindrome(string s) {
int len = s.length();
if (len <= 1) {
return s;
}
reset(len);
int max = 1;
int cur = 1;
int max_i = 0;
int max_j = 0;
for (int i = len - 1; i >= 0; i--) {
for (int j = i + 1; j < len; j++) {
if (( i+1 >= j-1 || dp[i+1][j-1]) && s[i] == s[j]) {
dp[i][j] = true;
cur = j - i + 1;
if (cur > max) {
max = cur;
max_i = i;
max_j = j;
}
}
}
}
return s.substr(max_i, max_j - max_i + 1);
}
private:
bool dp[1000][1000];
};
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