5. Longest Palindromic Substring
Posted 积少成多
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Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of Sis 1000, and there exists one unique longest palindromic substring.
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class Solution { public: string longestPalindrome(string s) { /**bool dp[i][j]代表s[i-j]是否回文 dp[i][j] = true ;i==j = s[i]==s[j] ;j=i+1 = (s[i]==s[j])&& dp[i+1][j-1]; j>i+1 要记住,dp[i][j]表示的i-j之间的字符串是不是回文, 当i==j时,dp[i][i]当然是回文 当j=i+1时,dp[i][j],要看字符串s[i]和s[j]之间是不是相同的? 当j>i+1时,dp[i][j]的判定情况,s[i]s[j]和dp[i+1][j-1]共同决定的; 怎么开始的? 外层循环j控制, 内存循环i判断当前能到达的位置是否是回文; */ int n = s.size(); bool dp[n][n]; fill_n(&dp[0][0],n*n,false);///初始化 int max_len = 1; int start = 0; for(int j = 0;j<s.size();j++){ for(int i = 0;i<=j;i++){ if(i==j) dp[i][j] = true; else if(j==(i+1)) dp[i][j] = s[i]==s[j]?true:false; else if(j>(i+1)) dp[i][j] = s[i]==s[j]&&dp[i+1][j-1]; if(dp[i][j]&&max_len < (j-i+1)){ max_len = j-i+1; start = i; } }///for } return s.substr(start,max_len); } };
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