LeetCode 第五题 Longest Palindromic Substring

先贴题目

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”

就是要找一个字符串里的回文串

第一种思路先写一个方法判断字符串是否是回文串，如果不是再将子串迭代一下，但这种情况的话”babadada”，如果第二个子串也是回文串则无法获取

class Solution {
    public String longestPalindrome(String s) {
        int lastindex=0;
        String str = null;
        for (int i=0;i<s.length();i++){
            lastindex=s.lastIndexOf(s.charAt(i));
            if(lastindex!=-1&&i!=lastindex){
                //说明存在
                str = s.substring(i,lastindex+1);
                if(palindrome(str)){
                    //表明是回文串
                    return str; 
                }else{
                    do{
                        //字符串中的字符串再搜一下
                        lastindex= s.lastIndexOf(s.charAt(i),s.length()-2);
                        if(lastindex!=-1){
                            str = s.substring(i,lastindex+1);
                            return longestPalindrome(str);
                        }
                    }while(s.lastIndexOf(s.charAt(i))==i);
                }
            }
    }
   return s.substring(0,1);
}
    //判断是否回文串
    private boolean palindrome(String s){
        for (int i=0;i<s.length()/2;i++){
            if(s.toCharArray()[i]!=s.toCharArray()[s.length()-i-1]){
                return false;
            }
        }
        return true;
    }
}

加了一个最大判断，当resultSize最大时获取result。”babadada”这个算是通过了。但是”aaabaaaa”又不行了。一看在else里面迭代的时候resultSize会重置，之前的resultSize就取不到了。

class Solution {
    public String longestPalindrome(String s) {
        int lastindex=0;
        String str = null;
        String result = s.substring(0,1);
        int resultSize = 0;
        for (int i=0;i<s.length();i++){
            lastindex=s.lastIndexOf(s.charAt(i));
            if(lastindex!=-1&&i!=lastindex){
                //说明存在
                str = s.substring(i,lastindex+1);
                if(palindrome(str)){
                    //表明是回文串
                    if(str.length()>resultSize){
                        result=str;
                        resultSize = str.length();
                    }
                }else{
                    do{
                        //字符串中的字符串再搜一下
                        lastindex= s.lastIndexOf(s.charAt(i),s.length()-2);
                        if(lastindex!=-1){
                            str = s.substring(i,lastindex+1);
                            longestPalindrome(str);
                        }
                    }while(s.lastIndexOf(s.charAt(i))==i);
                }
            }
    }
   return result;
}
    //判断是否回文串
    private boolean palindrome(String s){
        for (int i=0;i<s.length()/2;i++){
            if(s.toCharArray()[i]!=s.toCharArray()[s.length()-i-1]){
                return false;
            }
        }
        return true;
    }
}

又加了一个子方法，迭代子方法。终于可以通过了。

class Solution {
    public String longestPalindrome(String s) {

        String result = s.substring(0,1);
        int resultSize = 0;
        return longpal(s,resultSize,result);
}

    private String longpal(String s,int resultSize,String result){
               String str = null;
        int lastindex=0;
        for (int i=0;i<s.length();i++){
            lastindex = s.lastIndexOf(s.charAt(i));
            if (lastindex != -1 && i != lastindex) {
                //说明存在
                str = s.substring(i, lastindex + 1);
                if (palindrome(str)) {
                    //表明是回文串
                    if (str.length() > resultSize) {
                        result = str;
                        resultSize = str.length();
                    }
                } else {
                    //字符串长度-1去掉相同的重新查找
                    str = s.substring(0, s.length() - 1);
                    lastindex = str.lastIndexOf(s.charAt(i));
                    if (lastindex != -1 && i != lastindex) {
                        str = str.substring(i, lastindex + 1);
                        str=longpal(str,resultSize,result);
                        if (str.length() > resultSize) {
                            result = str;
                            resultSize = str.length();
                        }
                    }
                }
            }

        }
        return result;
    }
    //判断是否回文串
    private boolean palindrome(String s){
        for (int i=0;i<s.length()/2;i++){
            if(s.toCharArray()[i]!=s.toCharArray()[s.length()-i-1]){
                return false;
            }
        }
        return true;
    }
}

但是又给了我下面这个字符串让我找子串，我的方法耗时太长！@#￥%，你nb。想不出来了 看下答案

"civilwartestingwhetherthatnaptionoranynartionsoconceivedandsodedicatedcanlongendureWeareqmetonagreatbattlefiemldoftzhatwarWehavecometodedicpateaportionofthatfieldasafinalrestingplaceforthosewhoheregavetheirlivesthatthatnationmightliveItisaltogetherfangandproperthatweshoulddothisButinalargersensewecannotdedicatewecannotconsecratewecannothallowthisgroundThebravelmenlivinganddeadwhostruggledherehaveconsecrateditfaraboveourpoorponwertoaddordetractTgheworldadswfilllittlenotlenorlongrememberwhatwesayherebutitcanneverforgetwhattheydidhereItisforusthelivingrathertobededicatedheretotheulnfinishedworkwhichtheywhofoughtherehavethusfarsonoblyadvancedItisratherforustobeherededicatedtothegreattdafskremainingbeforeusthatfromthesehonoreddeadwetakeincreaseddevotiontothatcauseforwhichtheygavethelastpfullmeasureofdevotionthatweherehighlyresolvethatthesedeadshallnothavediedinvainthatthisnationunsderGodshallhaveanewbirthoffreedomandthatgovernmentofthepeoplebythepeopleforthepeopleshallnotperishfromtheearth"

正确答案的思路与我不同，我是从两头向中间判断，答案是从中间向两头判断。这样的话就不用判断子串是否符合条件 不需要迭代 减少了时间复杂度。

public String longestPalindrome(String s) {
    int start = 0, end = 0;
    for (int i = 0; i < s.length(); i++) {
        int len1 = expandAroundCenter(s, i, i);
        int len2 = expandAroundCenter(s, i, i + 1);
        int len = Math.max(len1, len2);
        if (len > end - start) {
            start = i - (len - 1) / 2;
            end = i + len / 2;
        }
    }
    return s.substring(start, end + 1);
}

private int expandAroundCenter(String s, int left, int right) {
    int L = left, R = right;
    while (L >= 0 && R < s.length() && s.charAt(L) == s.charAt(R)) {
        L--;
        R++;
    }
    return R - L - 1;
}

答案的时间复杂度为O(n​2​​).空间复杂度为O(1)
判断两种情况如“abba”,”abbcbba”比较大小。