[LeetCode] Longest Palindromic Substring 解题报告
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.
» Solve this problem
[解题思路]
O(n*n)。对于每一个字符，以之作为中间元素往左右寻找。注意处理奇偶两种模式：
1. aba
2. abba
1: string longestPalindrome(string s) {
2: // Start typing your C/C++ solution below
3: // DO NOT write int main() function
4: int startIndex = 0;
5: int len = 0;
6: int sI,eI;
7: for(int i =0; i< s.size() -1; i++)
8: {
9: if(s[i] == s[i+1])
10: {
11: sI = i;
12: eI = i+1;
13: Search(s, sI, eI, len, startIndex);
14: }
15: sI = i;
16: eI = i;
17: Search(s, sI, eI, len, startIndex);
18: }
19: if(len == 0)
20: len = s.size();
21: return s.substr(startIndex, len);
22: }
23: void Search(string&s, int sI, int eI, int&len, int& startIndex)
24: {
25: int step = 1;
26: while((sI-step)>=0&& (eI+step)<s.size())
27: {
28: if(s[sI-step] != s[eI+step])
29: {
30: break;
31: }
32: step++;
33: }
34: int wid = eI- sI+2*step -1;
35: if(wid > len)
36: {
37: len = wid;
38: startIndex = sI - step+1;
39: }
40: }
此题也有O(n)解法。相当巧妙， http://www.leetcode.com/2011/11/longest-palindromic-substring-part-ii.html
Version2: add DP solution, 3/5/2013
定义函数
P[i,j] = 字符串区间[i,j]是否为palindrome.
首先找个例子，比如S="abccb",
S= a b c c b
Index = 0 1 2 3 4
P[0,0] =1 //each char is a palindrome
P[0,1] =S[0] == S[1] , P[1,1] =1
P[0,2] = S[0] == S[2] && P[1,1], P[1,2] = S[1] == S[2] , P[2,2] = 1
P[0,3] = S[0] == S[3] && P[1,2], P[1,3] = S[1] == S[3] && P[2,2] , P[2,3] =S[2] ==S[3], P[3,3]=1
......................
由此就可以推导出规律
P[i,j] = 1 if i ==j
= S[i] ==S[j] if j = i+1
= S[i] == S[j] && P[i+1][j-1] if j>i+1
实现如下：
1: string longestPalindrome(string s) {
2: int len = s.size();
3: int P[len][len];
4: memset(P, 0, len*len*sizeof(int));
5: int maxL=0, start=0, end=0;
6: for(int i =0; i< s.size(); i++)
7: {
8: for(int j =0; j<i; j++)
9: {
10: P[j][i] = (s[j] == s[i] && (i-j<2 || P[j+1][i-1]));
11: if(P[j][i] && maxL < (i-j+1))
12: {
13: maxL = i-j+1;
14: start = j;
15: end = i;
16: }
17: }
18: P[i][i] =1;
19: }
20: return s.substr(start, end-start +1);
21: }