5. Longest Palindromic Substring LeetCode题解

Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.

题意：

给定一个字符串s，找出其中最长的回文子串，假定字符串最长为1000

Example:

Input: "babad"

Output: "bab"

Note: "aba" is also a valid answer.

Example:

Input: "cbbd"

Output: "bb"

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题解：

详见：http://blog.csdn.net/baidu_23318869/article/details/48154899

Code【Java】

public class Solution {
		// 在原字符串中插入指定字符
		public String encode(String s, String ch) {
			StringBuilder ss = new StringBuilder();
			for (int i = 0; i < s.length(); ++i) {
				ss.append(ch);
				ss.append(s.charAt(i));
			}
			ss.append(ch);
			return ss.toString();
		}
		// 删除原字符串中指定字符
		public String decode(String s, String ch) {
			return s.replaceAll( ch, "");
		}
		// 返回以当前pos为中心回文长度的一半（不含pos）
		int getHalfPalindromeLen(String s, int pos) {
			int i = 1;
			for ( ; 0 <= pos - i && pos + i < s.length(); ++i) {
				if (s.charAt(pos - i) != s.charAt(pos + i)) {
					break;
				}
			}
			return i - 1;
		}

		public String longestPalindrome(String s) {
			// 初始化字符串以及记录回文长度的数组
		    s = encode(s, "#");
		    int[] halfLen = new int[s.length()];
		    // 上一次最长回文串位置
		    int lastPos = 0;
		    int lastLen = 0;
		    for (int i = 0; i < s.length(); ++i) {
		    	if (i <= lastPos + lastLen) {
		    		int dis = i - lastPos;
		    		int cor = lastPos - dis;
		    		halfLen[i] = (cor - halfLen[cor] <= lastPos - lastLen) ?
		    					 getHalfPalindromeLen(s, i) : halfLen[cor];
		    	}
		    	else {
		    		halfLen[i] = getHalfPalindromeLen(s, i);
		    	}
		    	if (lastPos + lastLen < i + halfLen[i]) {
		    		lastPos = i;
		    		lastLen = halfLen[i];
		    	}
		    }
		    // 寻找最长回文串并返回
		    int maxPos = 0;
		    for (int i = 1; i < halfLen.length; ++i) {
		    	if (halfLen[maxPos] < halfLen[i]) {
		    		maxPos = i;
		    	}
		    }
		    return decode(s.substring(maxPos - halfLen[maxPos], maxPos + halfLen[maxPos] + 1), "#");
		}
		
	}

Code【C++】

class Solution {
	// 在原字符串中插入指定字符
	string encode(string s, char ch = '#') {
		stringstream ss;
		for (int i = 0; i < s.size(); ++i) {
			ss << ch << s[i];
		}
		ss << ch;
		return ss.str();
	}
	// 删除原字符串中指定字符
	string decode(string s, char ch = '#') {
		stringstream ss;
		for (int i = 0; i < s.size(); ++i) {
			if (s[i] != ch) {
				ss << s[i];
			}
		}
		return ss.str();
	}
	// 返回以当前pos为中心回文长度的一半（不含pos）
	int getHalfPalindromeLen(string s, int pos) {
		int i = 1;
		for ( ; 0 <= pos - i && pos + i < s.size(); ++i) {
			if (s[pos - i] != s[pos + i]) {
				break;
			}
		}
		return i - 1;
	}
public:
    string longestPalindrome(string s) {
    	// 初始化字符串 以及记录最长回文长度数组
    	s = encode(s);
    	vector<int> halfLen(s.size(), 0);

    	// 开始计算最长回文
    	int lastPos = 0;
    	int lastLen = 0;
    	for (int i = 0; i < s.size(); ++i) {
    		if (i <= lastPos + lastLen) {
    			int dis = i - lastPos;
    			int cor = lastPos - dis;
    			halfLen[i] = (cor - halfLen[cor] <= lastPos - lastLen) ?
    						 getHalfPalindromeLen(s, i) : halfLen[cor];
    		}
    		else {
    			halfLen[i] = getHalfPalindromeLen(s, i);
    		}
    		// 更新lastPos以及lastLen
    		if (lastPos + lastLen < i + halfLen[i]) {
    			lastPos = i;
    			lastLen = halfLen[i];
    		}
    	}
    	// 寻找最长回文位置
    	int maxPos = 0;
    	for (int i = 1; i < s.size(); ++i) {
    		if (halfLen[maxPos] < halfLen[i]) {
    			maxPos = i;
    		}
    	}
        return decode(s.substr(maxPos - halfLen[maxPos], halfLen[maxPos] * 2 + 1));
    }
};