letcode刷题--4.最长回文字符串

给你一个字符串 s，找到 s 中最长的回文子串。

方法一：暴力解法

超出时间限制

class Solution {
    public String longestPalindrome(String s) {
        String ans = "";
        int len = s.length();
        int max = 0;

        for (int i = 0; i < len; i ++){
            for (int j = i + 1; j <= len; j ++){
                String test = s.substring(i,j);
                if (isPalindrome(test) && test.length() > max){
                    ans = s.substring(i,j);
                    max = Math.max(max, ans.length());
                }
            }
        }
        return ans;
    }

    public boolean isPalindrome(String s){
        int len = s.length();
        for (int i = 0;i < len/2; i ++) {
            if (s.charAt(i) != s.charAt(len -i - 1)) {
                return false;
            }
        }
        return true;
    }
}

方法二：动态规划，相当于填写一个二维表，需要使用到状态转移方程

dp[i][j]的值需要计算dp[i+1][j-1]是不是true，以列来填表

class Solution {
    public String longestPalindrome(String s) {
        int len = s.length();
        if (len < 2) {
            return s;
        }
        int maxLen = 1;
        char[] chars = s.toCharArray();
        int begin = 0;
        boolean[][] dp = new boolean[len][len];
        for (int i = 0; i < len; i++) {
            dp[i][i] = true;
        }

        for (int j = 1; j < len; j++) {
            for (int i = 0; i < j; i++) {
                if (chars[i] != chars[j]) {
                    dp[i][j] = false;
                } else {
                    if (j - i < 3) {
                        dp[i][j] = true;
                    } else {
                        dp[i][j] = dp[i + 1][j - 1];
                    }
                }
                if (dp[i][j] && j - i + 1 > maxLen) {
                    maxLen = j - i + 1;
                    begin = i;
                }
            }
        }
        return s.substring(begin, begin + maxLen);
    }
}

方法三：中心扩散

class Solution {
    public String longestPalindrome(String s) {
        int len = s.length();
        int start = 0;
        int end = 0;
        int maxLen = 0;
        if (len<1 || s == null){
            return "";
        }
        else {
            for (int i = 0;i < len; i ++){
                int max1 = numsPalindrome(s,i,i);
                int max2 = numsPalindrome(s,i,i+1);
                maxLen = Math.max(max1,max2);
                if (maxLen > end - start) {
                    start = i - (maxLen - 1) / 2;
                    end = i + maxLen / 2;
                }
            }
            return s.substring(start,end+1);
        }
    }

    public int numsPalindrome(String s,int left,int right){
        while (left>=0 && right < s.length() && s.charAt(left) == s.charAt(right)){
            --left;
            ++right;
        }
        return right - left - 1;
    }
}