标签：note put val solution find NPU lse output return
Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.
Example 1:
Input: "babad" Output: "bab" Note: "aba" is also a valid answer.
Example 2:
Input: "cbbd" Output: "bb"
注意不存在回文串的处理。此题还有DP和马拉车解法。
1 class Solution { 2 public: 3 string longestPalindrome(string s) { 4 int resNum = 0; 5 string resStr = ""; 6 for(int i=0;i<s.size();i++){ 7 if(s[i] == s[i+1]&&i+1<s.size()){ //偶数 8 int j; 9 for(j=1;i-j>=0&&i+1+j<s.size();j++){ 10 if(s[i-j] != s[i+1+j]) break; 11 } 12 if(2*j>resNum) { 13 resNum = 2*j; 14 resStr = s.substr(i-j+1,2*j); 15 } 16 } 17 if(s[i] == s[i+2]&&i+2<s.size()){ //奇数 18 int j; 19 for(j=1;i-j>=0&&i+2+j<s.size();j++){ 20 if(s[i-j] != s[i+2+j]) break; 21 } 22 if(2*j+1>resNum) { 23 resNum = 2*j+1; 24 resStr = s.substr(i-j+1,2*j+1); 25 } 26 } 27 } 28 if(resStr == ""&&s.size()>0) resStr = s.substr(0,1); 29 return resStr; 30 } 31 };
DP的做法，递推公式：
设dp[i][j]为i到j的子串是否为回文，则有
1.dp[i][j] = true （i == j）
2.dp[i][j] = s[i]==s[j] （i+1 == j）
3.dp[i][j] = s[i]==s[j]&&dp[i+1][j-1] (i+1<j)
注意遍历的顺序，依赖关系，可以从左向右，或者从上到下遍历。
从左到右：
1 class Solution { 2 public: 3 string longestPalindrome(string s) { 4 int dp[1000+1][1000+1] = {0}, left = 0, right = 0, len = 0; 5 for (int i = 0; i < s.size(); ++i) { 6 for (int j = 0; j < i; ++j) { 7 if(j+1 == i) dp[j][i] = s[j]==s[i]?1:0; 8 else dp[j][i] = s[j]==s[i]&&dp[j+1][i-1]?1:0; 9 if(dp[j][i] && i-j+1>len){ 10 len = i-j+1; 11 left = j; 12 right = i; 13 } 14 } 15 dp[i][i] = 1; 16 } 17 return s.substr(left, right - left + 1); 18 } 19 };
从上到下：
1 class Solution { 2 public: 3 string longestPalindrome(string s) { 4 int dp[1000+1][1000+1] = {0}, left = 0, right = 0, len = 0; 5 for (int i = s.size()-1; i >= 0; i--) { 6 for (int j = s.size()-1; j > i; j--) { 7 if(i+1 == j) dp[i][j] = s[j]==s[i]?1:0; 8 else dp[i][j] = s[j]==s[i]&&dp[i+1][j-1]?1:0; 9 if(dp[i][j] && j-i+1>len){ 10 len = j-i+1; 11 left = i; 12 right = j; 13 } 14 } 15 dp[i][i] = 1; 16 } 17 return s.substr(left, right - left + 1); 18 } 19 };
【leetcode】5. Longest Palindromic Substring
标签：note put val solution find NPU lse output return
原文地址：https://www.cnblogs.com/yuchi328/p/9692844.html
