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-优快云博客

本文介绍了一种用于寻找字符串中最长对称子串的算法实现。通过示例代码展示了如何针对不同输入字符串找到其最长对称子串的长度，并解释了算法背后的逻辑。

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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;
	}
}