Longest Palindromic Substring
思路:这道题采用遍历法,分两种情况考虑;
1.ada类型
2.aadd类型
对于上述两种类型,分别以k、k+1索引 或者k、k+2索引来判断,然后设置两个标记 i、j 索引,分别向两边计数,当出现j-i有最大值时,返回该索引。
代码如下:
public class Solution {
public String longestPalindrome(String s) {
int length =s.length();
int max1=0,max2=0;//记录两个索引
int len = 0;
if(length == 2 && s.charAt(0)==s.charAt(1)||length ==1){
return s;
}
for (int k = 0; k < length-1; k++) {
if(s.charAt(k)==s.charAt(k+1)){
if(len<=1){
max1 = k;
max2 =k+1;
len = 1;
}
int i = k-1;
int j = k+2;
while(i>=0&&j<=length-1){
if(s.charAt(i)==s.charAt(j)){
if(j-i>len){
max1 = i;
max2 = j;
len = max2 - max1;
}
i--;
j++;
}else{
break;
}
}
}
if(k+2>=length){
continue;
}
if (s.charAt(k)==s.charAt(k+2)) {
if(len<=2){
max1 = k;
max2 =k+2;
len = 2;
}
int i = k-1;
int j = k+3;
while(i>=0&&j<=length-1){
if(s.charAt(i)==s.charAt(j)){
if(j-i>len){
max1 = i;
max2 = j;
len = max2 - max1;
}
i--;
j++;
}else{
break;
}
}
}
}
return s.substring(max1, max2+1);
}
}
这里有个优化算法,值得一看
public class Solution {
public String longestPalindrome(String s) {
int length =s.length();
int max1=0,max2=0;//记录两个索引
int len = 0;
if(length == 2 && s.charAt(0)==s.charAt(1)||length ==1){
return s;
}
for (int k = 0; k < length-1; k++) {
if(s.charAt(k)==s.charAt(k+1)){
if(len<=1){
max1 = k;
max2 =k+1;
len = 1;
}
int i = k-1;
int j = k+2;
while(i>=0&&j<=length-1){
if(s.charAt(i)==s.charAt(j)){
if(j-i>len){
max1 = i;
max2 = j;
len = max2 - max1;
}
i--;
j++;
}else{
break;
}
}
}
if(k+2>=length){
continue;
}
if (s.charAt(k)==s.charAt(k+2)) {
if(len<=2){
max1 = k;
max2 =k+2;
len = 2;
}
int i = k-1;
int j = k+3;
while(i>=0&&j<=length-1){
if(s.charAt(i)==s.charAt(j)){
if(j-i>len){
max1 = i;
max2 = j;
len = max2 - max1;
}
i--;
j++;
}else{
break;
}
}
}
}
return s.substring(max1, max2+1);
}
}