LeetCode Shortest Palindrome(kmp的妙用)

题意：给出一个字符串，通过在字符串前添加字符，使得新的字符串为最短的对称字符串

思路：假设原字符串为s,将其取反的字符串为text,求text的后缀与s的前缀相同时的索引，将text与s索引后的字符拼接起来就是要求的结果

具体代码如下：

public class Solution
{private int[] kmp_table(String s){int[] next = new int[s.length()];next[0] = -1;int j = 0;for (int i = 1; i < s.length(); i++) {int k = next[i - 1];while (k > -1 && s.charAt(k) != s.charAt(i - 1)) {k = next[k];}next[i] = k + 1;}return next;}private int kmp_search(String text, String pattern){int[] next = kmp_table(pattern);int m = 0, i = 0;while (m + i < text.length()) {while (pattern.charAt(i) == text.charAt(m + i)) {if (m + i == text.length() - 1) {return m;}i++;}m += i - next[i];i = next[i] > -1 ? next[i] : 0;}return 0;}public String shortestPalindrome(String s){if (s.isEmpty()) return s;String text = new StringBuilder(s).reverse().toString();int index = kmp_search(text, s);StringBuilder ans = new StringBuilder();ans.append(text);ans.append(s.substring(text.length() - index));return ans.toString();}
}

解法二：

public class Solution {private int[] kmp_table(String s){int[] next = new int[s.length()];next[0] = -1;int j = -1, i = 0;while (i < s.length()){while (j > -1 && s.charAt(i) != s.charAt(j)){j = next[j];}i++;j++;if (i < s.length() && j < s.length()){if (s.charAt(i) == s.charAt(j)) next[i] = next[j];else{next[i] = j;}}}return next;}private int kmp_search(String text, String pattern){int[] next = kmp_table(pattern);int m = 0, i = 0;while (m < text.length()) {while (i > -1 && pattern.charAt(i) != text.charAt(m)) {i = next[i];}m++;i++;if (i >= pattern.length()){break;}}return m - i;}public String shortestPalindrome(String s){if (s.isEmpty()) return s;String text = new StringBuilder(s).reverse().toString();int index = kmp_search(text, s);StringBuilder ans = new StringBuilder();ans.append(text);ans.append(s.substring(text.length() - index));return ans.toString();}
}

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