java-73-输入一个字符串，输出该字符串中对称的子字符串的最大长度

public class LongestSymmtricalLength {

	/*
	 * Q75题目：输入一个字符串，输出该字符串中对称的子字符串的最大长度。
	 * 比如输入字符串“google”，由于该字符串里最长的对称子字符串是“goog”，因此输出4。
	 */
	
	public static void main(String[] args) {
		String[] strs={
				"google",
				"elgoog",
				"agollloge",
				"aba",
				"aaaaaa",
		};
		for(String each:strs){
			int len=longestSymmtricalLength(each);
			System.out.println(len);
		}
	}

	/*
	 * consider this string:
	 * ..aAa..(A represents a string of any length)
	 * if we already know 'A' is a palindrome,
	 * we just need to test if the char before 'A' and that after 'A' are the same
	 * if so,A=aAa now.
	 * Do it again and again to find the result.
	 */
	public static int longestSymmtricalLength(String str){
		if(str==null||str.length()==0){
			return -1;
		}
		int symLen=1;
		char[] letter=str.toCharArray();
		int strLen=str.length();
		int curIndex=1;
		while(curIndex>0&&curIndex<strLen-1){
			//odd symmetrical length,the 'pivot' char is letter[curIndex]
			int i=curIndex-1;
			int j=curIndex+1;
			while(i>=0&&j<=(strLen-1)&&letter[i]==letter[j]){
				i--;
				j++;
			}
			int newLen=j-i-1;
			if(newLen>symLen){
				symLen=newLen;
			}
			//even symmetrical length,the 'pivot' chars are letter[curIndex] and letter[curIndex+1]
			i=curIndex;
			j=curIndex+1;
			while(i>=0&&j<=(strLen-1)&&letter[i]==letter[j]){
				i--;
				j++;
			}
			newLen=j-i-1;
			if(newLen>symLen){
				symLen=newLen;
			}
			curIndex++;
		}
		return symLen;
	}
}