lc_模拟_回文串_0921

lc5 最长回文子串

1 动态规划

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

2 中心扩展算法

class Solution {
    public String longestPalindrome(String s) {
        int length = s.length();
        if (length < 2) {
            return s;
        }
        int start = 0, end = 0;
        for (int i = 0; i < length; i++) {
            int len1 = expand(s, i, i);
            int len2 = expand(s, i, i + 1);
            int len = Math.max(len1, len2);
            if (len > end - start) {
// 合并了为奇数偶数的两种情况的计算
                start = i - (len - 1) / 2;
                end = i + len / 2;
            }
        }
        return s.substring(start, end + 1);
    }
    public int expand(String s, int left, int right) {
        while (left >= 0 && right < s.length() && s.charAt(left) == s.charAt(right)) {
            --left;
            ++right;
        }
        return right - left - 1; 
    }
}

3 Manacher

略

lc🗡0027 回文链表

利用递归

class Solution {
    private ListNode frontPointer;
    public boolean isPalindrome(ListNode head) {
        frontPointer = head;
        return recursivelyCheck(head);
    }
    private boolean recursivelyCheck(ListNode curr) {
        if (curr != null) {
            if (!recursivelyCheck(curr.next)) {
                return false;
            }
            if (curr.val != frontPointer.val) {
                return false;
            }
            frontPointer = frontPointer.next;
        }
        return true;
    }
}

时间复杂度O(n),空间复杂度O(n);方法很取巧

快慢指针加反转链表

class Solution {
    public boolean isPalindrome(ListNode head) {
        if (head == null) {
            return true;
        }
        ListNode firstHalfEnd = endOfFirstHalf(head);
        ListNode secondHalfStart = reverseList(firstHalfEnd.next);

        ListNode p1 = head;
        ListNode p2 = secondHalfStart;
        boolean res = true;
        while (res && p2 != null) {
            if (p1.val != p2.val) {
                res = false;
            }
            p1 = p1.next;
            p2 = p2.next;
        }
        firstHalfEnd.next = reverseList(secondHalfStart);
        return res;
    }
    private ListNode reverseList(ListNode head) {
        ListNode prev = null;
        ListNode curr = head;
        while (curr != null) {
            ListNode next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        return prev;
    }

    private ListNode endOfFirstHalf(ListNode head) {
        ListNode fast = head;
        ListNode slow = head;
        while (fast.next != null && fast.next.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        return slow;
    }
}