leetcode5

给你一个字符串 s，找到 s 中最长的 

回文串

示例 1：

输入：s = "babad"
输出："bab"
解释："aba" 同样是符合题意的答案。

最开始我是将回文串都存入map。使用substr进行切割，i为起始，j-i+1是步长

string longestPalindrome(string s) {if (s.length() <= 1) {return s;}map<int,string> resultmap;string temp;for (int i = 0; i < s.length() - 1; i++) {for (int j = i; j < s.length(); j++) {               //cout << "temp:" << temp << endl;string s1 = s.substr(i, j-i+1);temp = s1;reverse(temp.begin(), temp.end());//cout << "s1:" << s1 << endl;if (temp == s1) {resultmap[i] = s1;//cout << "s1:" << s1 << endl;}}}int length = 0;int index = 0;for (int i = 0; i < resultmap.size(); i++) {if (resultmap[i].length() > length) {length = resultmap[i].length();index = i;}}return resultmap[index];
}

优化了一下，还是时间复杂度高

string longestPalindrome(string s) {if (s.length() <= 1) {return s;}//map<int,string> resultmap;string temp;int max = 0;string maxstr;for (int i = 0; i < s.length() - 1; i++) {for (int j = i; j < s.length(); j++) {                             string s1 = s.substr(i, j - i + 1);temp = s1;if (s1.length() > max){reverse(temp.begin(), temp.end());//cout << "s1:" << s1 << endl;//cout << "temp:" << temp << endl;if (temp == s1) {maxstr = s1;max = s1.length();//cout << "s1:" << s1 << endl;}}               }}return maxstr;
}

最终还是用动态规划的方式解决。动态规划的核心思想把已经发生过的情况储存起来，需要时直接调用，也就是空间换时间。

dp存储了某一段字符串是否是回文串，dp[i][j]的值如果为true，表示字符串s中从下标i到下标j的子串是一个回文串；如果为false，则不是回文串。

动态规划需要递推，所以也需要初始化，这里需要初始化单个字符串与连续字符串的情况，之后递推。

string longestPalindrome(string s) {int n = s.length();if (n <= 1) {return s;}vector <vector<bool>> dp(n, vector<bool>(n, false));for (int i = 0; i < s.length(); i++) {dp[i][i] = true;}for (int i = 0; i < s.length()-1; i++) {if (s[i]==s[i+1]){dp[i][i+1] = true;dp[i+1][i] = true;}}int index=0;int maxlen=0;for (int len = 1; len < s.length(); len++) {for (int j = 0; j + len < s.length(); j++) {//cout << s[j] << endl << s[j + len]<<endl;if (s[j] == s[j + len] && dp[j + 1][j + len - 1]==true) {dp[j][j + len] = true;dp[j+len][j] = true;index = j;maxlen = len;cout << "找到回文" << ":";cout << index << "," << maxlen;}}}string res = s.substr(index, maxlen+1);return res;
}