public class LongestSymmtricalLength2 {	/*	 * 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 = { "a","google", "elgoog", "agollloge", "aba", "aaaaaa", };		for (String each : strs) {			int len = longestSymmtricalLength(each);			System.out.println("l1:" + len);			len = longestSymmtricalLength2(each);			System.out.println("l2:" + len);		}	}	public static int longestSymmtricalLength(String str) {		if (str == null || str.length() == 0) {			return -1;		}		//去掉aa为1实际为2 的bug		if(str.length()==2){			if(str.charAt(0)==str.charAt(1))				return 2;		}		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] goog  就是用位置 1和2进行对称			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;	}	/*	 * 上面的方法是从位置1,进行向外扩大半径进行搜索,比如google,currIndex=1;i=0;j=1;(因为对称相等就是回文)，向外扩大半径搜索	 * 改版之后的算法,不在区分奇偶	 */	public static int longestSymmtricalLength2(String str) {		if (str == null || str.length() == 0) {			return -1;		}		StringBuffer buffer = new StringBuffer();		buffer.append("#");		for (int i = 0; i < str.length(); i++) {			buffer.append(str.charAt(i) + "#");		}		str = buffer.toString();		buffer.delete(0, buffer.length());		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) / 2;			if (newLen > symLen) {				symLen = newLen;			}			curIndex++;		}		return symLen;	}}