leetcode[5]Longest Palindromic Substring(最長迴文子串)

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

Example:

Input: "babad"

Output: "bab"
Note: "aba" is also a valid answer.

Example:

Input: "cbbd"
Output: "bb"

---

Solution1(DP)

public class Solution {

public String longestPalindrome(String s) {

//DP

if(s == null || s.isEmpty()){

return "";

}

int n = s.length() ;

int maxI = 0 ;

int maxJ = 0 ;

boolean [][] isPa = new boolean [n][n];

for(int i = 0 ; i < n ; i ++){

isPa [i][i] = true ;

if((i + 1 ) < n && s.charAt(i) == s.charAt(i + 1)){

isPa[i][i + 1] = true ;

maxI = i ;

maxJ = i + 1 ;

}

}

    //assume l >= 3
    for(int l = 3 ; l <= n ; l ++){
        for(int i = 0 ; i < n - l + 1 ; i ++){//i < n - l + 1
            int j  = i + l - 1;//pay attention to array's bound
            isPa[i][j] = (isPa[i + 1 ][j - 1] && s.charAt(i) == s.charAt(j));//字符串[i,j]是迴文的當且僅當[i+1,j-1]是迴文的且字符i等於字符j
            if(isPa[i][j]){
                maxI = i ;
                maxJ = j ;
            }
        }
    }
    return s.substring(maxI , maxJ + 1);
}
//assume l >= 3
    for(int l = 3 ; l <= n ; l ++){
        for(int i = 0 ; i < n - l + 1 ; i ++){//i < n - l + 1
            int j  = i + l - 1;//pay attention to array's bound
            isPa[i][j] = (isPa[i + 1 ][j - 1] && s.charAt(i) == s.charAt(j));//字符串[i,j]是迴文的當且僅當[i+1,j-1]是迴文的且字符i等於字符j
            if(isPa[i][j]){
                maxI = i ;
                maxJ = j ;
            }
        }
    }
    return s.substring(maxI , maxJ + 1);
}}

Solution2:(往回文字符串的兩邊拓展尋找更長的迴文字符串)

public class Solution {

public String longestPalindrome(String s) {

if (s == null) {

return "";

}

char[] arr = s.toCharArray();

int max = 0;

int maxi = 0;

int maxj = 0;

    int len = arr.length;

    for (int i = 0; i < len;) {
        int i1 = getFarestSameElementIndex(arr, i,len);
        int dist = getDistance(arr, i, i1,len);
        int index1 = i - dist;
        int index2 = i1 + dist;
        int l = index2 - index1;
        if (l > max) {
            max = l;
            maxi = index1;
            maxj = index2;
        }
        i = i1 + 1;//jump over the same chars 
    }

    return s.substring(maxi, maxj + 1);
}

    //以[index1,index2]做爲已是迴文的字符串，想兩邊拓展，返回最大拓展距離
private int getDistance(char[] arr, int index1, int index2,int len) {
    int i1 = index1 - 1;
    int i2 = index2 + 1;
    int dist = 0;
    while (i1 >= 0 && i2 < len) {
        if (arr[i1] == arr[i2]) {
            dist++;
        } else {
            break;
        }
        i1--;
        i2++;
    }
    return dist;
}

    //返回index開始的相同字符的最大長度
private int getFarestSameElementIndex(char[] arr, int index,int len) {
    for (int i = index + 1; i < len; i++) {
        if (arr[i] != arr[index]) {
            return i - 1;
        }
    }
    return len - 1;
}
int len = arr.length;

    for (int i = 0; i < len;) {
        int i1 = getFarestSameElementIndex(arr, i,len);
        int dist = getDistance(arr, i, i1,len);
        int index1 = i - dist;
        int index2 = i1 + dist;
        int l = index2 - index1;
        if (l > max) {
            max = l;
            maxi = index1;
            maxj = index2;
        }
        i = i1 + 1;//jump over the same chars 
    }

    return s.substring(maxi, maxj + 1);
}

    //以[index1,index2]做爲已是迴文的字符串，想兩邊拓展，返回最大拓展距離
private int getDistance(char[] arr, int index1, int index2,int len) {
    int i1 = index1 - 1;
    int i2 = index2 + 1;
    int dist = 0;
    while (i1 >= 0 && i2 < len) {
        if (arr[i1] == arr[i2]) {
            dist++;
        } else {
            break;
        }
        i1--;
        i2++;
    }
    return dist;
}

    //返回index開始的相同字符的最大長度
private int getFarestSameElementIndex(char[] arr, int index,int len) {
    for (int i = index + 1; i < len; i++) {
        if (arr[i] != arr[index]) {
            return i - 1;
        }
    }
    return len - 1;
}}