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Copy pathMiddle_106_5_Longest Palindromic Substring.java
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Middle_106_5_Longest Palindromic Substring.java
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方法一:
public class Solution {
public String longestPalindrome(String s) {
char[] ca = s.toCharArray();
int rs = 0, re = 0;
int max = 0;
for(int i = 0; i < ca.length; i++) {
if(isPalindrome(ca, i - max - 1, i)) {
rs = i - max - 1; re = i;
max += 2;
} else if(isPalindrome(ca, i - max, i)) {
rs = i - max; re = i;
max += 1;
}
}
return s.substring(rs, re + 1);
}
private boolean isPalindrome(char[] ca, int s, int e) {
if(s < 0) return false;
while(s < e) {
if(ca[s++] != ca[e--]) return false;
}
return true;
}
}
方法二:(manacher算法,改编自wikipaedia上的代码)
public class Solution {
public String longestPalindrome(String s) {
if (s==null || s.length()==0)
return "";
char[] s2 = addBoundaries(s.toCharArray());
int[] p = new int[s2.length];
int c = 0, r = 0; // Here the first element in s2 has been processed.
int m = 0, n = 0; // The walking indices to compare if two elements are the same
for (int i = 1; i<s2.length; i++) {
if (i>r) {
p[i] = 0; m = i-1; n = i+1;
} else {
int i2 = c*2-i;
if (p[i2]<(r-i)) {
p[i] = p[i2];
m = -1; // This signals bypassing the while loop below.
} else {
p[i] = r-i;
n = r+1; m = i*2-n;
}
}
while (m>=0 && n<s2.length && s2[m]==s2[n]) {
p[i]++; m--; n++;
}
if ((i+p[i])>r) {
c = i; r = i+p[i];
}
}
int len = 0; c = 0;
for (int i = 1; i<s2.length; i++) {
if (len<p[i]) {
len = p[i]; c = i;
}
}
char[] ss = Arrays.copyOfRange(s2, c-len, c+len+1);
return String.valueOf(removeBoundaries(ss));
}
private char[] addBoundaries(char[] cs) {
if (cs==null || cs.length==0)
return "||".toCharArray();
char[] cs2 = new char[cs.length*2+1];
for (int i = 0; i<(cs2.length-1); i = i+2) {
cs2[i] = '|';
cs2[i+1] = cs[i/2];
}
cs2[cs2.length-1] = '|';
return cs2;
}
private char[] removeBoundaries(char[] cs) {
if (cs==null || cs.length<3)
return "".toCharArray();
char[] cs2 = new char[(cs.length-1)/2];
for (int i = 0; i<cs2.length; i++) {
cs2[i] = cs[i*2+1];
}
return cs2;
}
}