5. Longest Palindromic Substring
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<pre style="box-sizing: border-box; overflow: auto; font-family: Menlo, Monaco, Consolas, 'Courier New', monospace; font-size: 13px; padding: 9.5px; margin-top: 0px; margin-bottom: 10px; line-height: 1.42857; color: rgb(51, 51, 51); word-break: break-all; word-wrap: break-word; border: 1px solid rgb(204, 204, 204); border-top-left-radius: 4px; border-top-right-radius: 4px; border-bottom-right-radius: 4px; border-bottom-left-radius: 4px; font-variant-ligatures: normal; font-variant-position: normal; font-variant-numeric: normal; font-variant-alternates: normal; font-variant-east-asian: normal; widows: 1; background-color: rgb(245, 245, 245);">还有一种遍历中心的方法时间复杂度也为n方,还有一种为n的方法
class Solution {public: string longestPalindrome(string s) { int length = s.length(); if(length == 0 || length == 1) return s; vector<vector<int>> dp(length+5,vector<int>(length+5,0)); dp[length-1][length-1]=1; int start = 0,max = 0; for(int i=0;i<length-1;i++){ dp[i][i]=1; if(s[i]==s[i+1]) { dp[i][i+1]=1; start = i; max = 2; } } for(int j=2;j<length;j++){ for(int i=0;i<=j;i++){ if(s[i]==s[j]&&dp[i+1][j-1]){ if(j-i+1>max){ start = i; max = j-i+1; } dp[i][j]=1; } } } for(int i=0;i<length;i++){ for(int j=0;j<length;j++) cout<<dp[i][j]; cout<<endl; } return s.substr(start,max); }};
"abaas"
10100
01000
00110
00010
00001
0 0
- 5.Longest Palindromic Substring
- 5. Longest Palindromic Substring
- 5. Longest Palindromic Substring
- 5.Longest Palindromic Substring
- 5. Longest Palindromic Substring
- 5.Longest Palindromic Substring
- 5. Longest Palindromic Substring
- 5. Longest Palindromic Substring
- 5. Longest Palindromic Substring
- 5.Longest Palindromic Substring
- 5. Longest Palindromic Substring
- 5. Longest Palindromic Substring
- 5. Longest Palindromic Substring
- 5. Longest Palindromic Substring
- 5. Longest Palindromic Substring
- 5. Longest Palindromic Substring
- 5. Longest Palindromic Substring
- 5. Longest Palindromic Substring
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