字符串中最长回文，最笨的解法

Posted 2022-04-21 gavinyang

回文：aba abcba

双重循环遍历字符串，外层从第一个开始找，内层循环从最后一个开始找。当外层的字符和内存循环的字符相等时则组成新的数组，判断是否是回文

public static void main(String[] args) 
    String str = "gcgdecdabcdefgfeh";
    char[] chars = str.toCharArray();
    Map<String,Integer> maxLength = maxLength(chars);
    System.out.println(maxLength);

返回结果：fgf=3, gcg=3, efgfe=5

//复制出新数组
private static char[] copynew(char[] chars,int start,int end)
    char[] newchars = new char[end-start+1];
    int n = 0;
    for (int i=start;i<=end;i++)
        newchars[n++] = chars[i];
    
    return newchars;


//判断是否是回文
public static boolean check(char[] chars)

    int start = 0;
    int end = chars.length-1;

    while (start<end)

        if(chars[start]==chars[end])
            start++;
            end--;
        else 
            return false;
        
    
    return true;