最长子序列回文问题,Longest Palindromic Subsequence
2017-06-03 16:01
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Given a string s, find the longest palindromic subsequence's length in s. You may assume that the maximum length of s is 1000.
Example 1:
Input:
Output:
One possible longest palindromic subsequence is "bbbb".
Example 2:
Input:
Output:
One possible longest palindromic subsequence is "bb".
public class longestsubPalindrome {
public static int longsubPalin(String s) { //求最长回文子序列长度
char[] ch = s.toCharArray();
int[][] dp = new int[ch.length][ch.length];
for (int i = 0; i < ch.length; i++) {
dp[i][i] = 1;
}
for (int i = 1; i < ch.length; i++) {
for (int j = i - 1; j >= 0; j--) {
if (ch[i] == ch[j])
dp[j][i] = dp[j + 1][i - 1] + 2;
else {
dp[j][i] = Math.max(dp[j + 1][i], dp[j][i - 1]);
}
}
}
return dp[0][ch.length - 1];
}
public static String subPalin(String s) { //求最长子序列对应长度的字符串
char[] ch = s.toCharArray();
String[][] dp = new String[ch.length][ch.length];
for(int i=0;i<ch.length;i++){
for(int j=0;j<ch.length;j++){
dp[i][j]="";
}
}
for (int i = 0; i < ch.length; i++) {
dp[i][i] = dp[i][i]+ch[i];
}
for (int i = 1; i < ch.length; i++) {
for (int j = i - 1; j >= 0; j--) {
if (ch[i] == ch[j]) {
dp[j][i] = dp[j + 1][i - 1]+ch[j];
dp[j][i]=ch[j]+dp[j][i];
} else {
dp[j][i] = dp[j + 1][i].length()>dp[j][i - 1].length()?dp[j + 1][i]:dp[j][i - 1];
}
}
}
return dp[0][ch.length - 1].toString();
}
public static void main(String[] args) {
System.out.println(longsubPalin("cbbd"));
System.out.println(subPalin("cbbd"));
System.out.println(longsubPalin("bbabb"));
System.out.println(subPalin("bbabb"));
}
}
Example 1:
Input:
"bbbab"
Output:
4
One possible longest palindromic subsequence is "bbbb".
Example 2:
Input:
"cbbd"
Output:
2
One possible longest palindromic subsequence is "bb".
public class longestsubPalindrome {
public static int longsubPalin(String s) { //求最长回文子序列长度
char[] ch = s.toCharArray();
int[][] dp = new int[ch.length][ch.length];
for (int i = 0; i < ch.length; i++) {
dp[i][i] = 1;
}
for (int i = 1; i < ch.length; i++) {
for (int j = i - 1; j >= 0; j--) {
if (ch[i] == ch[j])
dp[j][i] = dp[j + 1][i - 1] + 2;
else {
dp[j][i] = Math.max(dp[j + 1][i], dp[j][i - 1]);
}
}
}
return dp[0][ch.length - 1];
}
public static String subPalin(String s) { //求最长子序列对应长度的字符串
char[] ch = s.toCharArray();
String[][] dp = new String[ch.length][ch.length];
for(int i=0;i<ch.length;i++){
for(int j=0;j<ch.length;j++){
dp[i][j]="";
}
}
for (int i = 0; i < ch.length; i++) {
dp[i][i] = dp[i][i]+ch[i];
}
for (int i = 1; i < ch.length; i++) {
for (int j = i - 1; j >= 0; j--) {
if (ch[i] == ch[j]) {
dp[j][i] = dp[j + 1][i - 1]+ch[j];
dp[j][i]=ch[j]+dp[j][i];
} else {
dp[j][i] = dp[j + 1][i].length()>dp[j][i - 1].length()?dp[j + 1][i]:dp[j][i - 1];
}
}
}
return dp[0][ch.length - 1].toString();
}
public static void main(String[] args) {
System.out.println(longsubPalin("cbbd"));
System.out.println(subPalin("cbbd"));
System.out.println(longsubPalin("bbabb"));
System.out.println(subPalin("bbabb"));
}
}
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