Longest Palindromic Substring 在一个字符串里找最长回文子串@LeetCode

题目是不难（思路好想），但要细心写才不会错！特别注意到有偶数个和奇数个palindrome的情况，以及两者的优先级不同！

对每个字符，看看它的左边和右边是不是相等（奇数）

或者它自己和右边是否相等（偶数）

另：我专门写一篇文章总结这道题：http://blog.csdn.net/fightforyourdream/article/details/21309759

package Level4;

/**
 * Longest Palindromic Substring
 * 
 * Given a string S, find the longest palindromic substring in S. You may assume
 * that the maximum length of S is 1000, and there exists one unique longest
 * palindromic substring.
 * 
 */
public class S5 {

	public static void main(String[] args) {
		String s = "abbbbda";
		System.out.println(longestPalindrome(s));
	}

	public static String longestPalindrome(String s) {
		if(s.length() == 0){
			return "";
		}
		int maxDiff = 0;
		int maxIndex = 0;
		int len = s.length();
		boolean odd = true;		// 默认是奇数个palindrome（即没有palindrome的情况也属于奇数）
		
		// 先测试palindrome是偶数情况，因为我默认"aa"和"a"都是diff为0的情况，这时palindrome为偶数的更长一些，所以给偶数更高优先级
		for(int i=0; i<len; i++){
			int diff = 0;
			if(i+1<len && s.charAt(i)==s.charAt(i+1)){		// Even
				diff = 0;
				while(i-diff>=0 && i+1+diff<len && s.charAt(i-diff)==s.charAt(i+1+diff)){
					diff++;
				}
				if(diff-1 >= maxDiff){		// 给偶数更高优先级（尽量找偶数个palindrome）
					maxDiff = diff-1;
					maxIndex = i;
					odd = false;
				}
			}

			// 再测试palindrome是奇数个情况，只有当奇数palindrome的diff超过（不能一样）偶数palindrome的diff时，才采纳
			diff = 0;			// Odd
			while(i-diff>=0 && i+diff<len && s.charAt(i-diff)==s.charAt(i+diff)){
				diff++;
			}
			if(diff-1 > maxDiff){
				maxDiff = diff-1;
				maxIndex = i;
				odd = true;
			}
			
		}
		
//		System.out.println("maxDiff:" + maxDiff);
		if(odd){		// 奇数个palindrome
//			System.out.println("odd");
			return s.substring(maxIndex-maxDiff, maxIndex+maxDiff+1);
		}else{		// 偶数个palindrome
//			System.out.println("even");
			return s.substring(maxIndex-maxDiff, maxIndex+maxDiff+2);
		}
		
	}

}

采用中心展开法：

public class Solution {
    public String longestPalindrome(String s) {
        if(s.length() == 0) {
            return s;
        }
        int max = 0;
        String maxs = s;
        for(int i=0; i<s.length(); i++) {
            int p = i;
            int dist = 0;
            while(p-dist >= 0 && p+dist < s.length() && s.charAt(p-dist) == s.charAt(p+dist)) {
                dist++;
            }
            dist--;
            int len = (i+dist) - (i-dist) + 1;
            if(len > max) {
                max = len;
                maxs = s.substring(i-dist, i+dist+1);
            }
            
            p = i;
            dist = 0;
            while(i-dist >= 0 && i+1+dist < s.length() && s.charAt(i-dist) == s.charAt(i+1+dist)) {
                dist++;
            }
            dist--;
            len = (i+1+dist) - (i-dist) + 1;
            if(len > max) {
                max = len;
                maxs = s.substring(i-dist, i+1+dist+1);
            }
        }
        
        return maxs;
    }
}