string match 字符串匹配

214. Shortest Palindrome Hard

You are given a string s. You can convert s to a palindrome by adding characters in front of it.

Return the shortest palindrome you can find by performing this transformation. 

Example 1:

Input: s = "aacecaaa"
Output: "aaacecaaa"

Example 2:

Input: s = "abcd"
Output: "dcbabcd" 

Constraints:

• 0 <= s.length <= 5 * 104
• s consists of lowercase English letters only.

思路：首先如何满足palindrom，从开始数，找到最长的palindrome，然后把剩余不是的部分倒转拼到左侧即实现了palindrome

解法1：暴力尝试所有情况，时间复杂读为O(N²)

解法一就不写了

解法2: 使用rolling hash，如果是palindrome的话，从左右两侧计算hash值应该是同一个值

因此，我们可以从第一个字母开始逐步向后计算hash值，一个维护从前到后的hash值，一个维护从后向前的hash值

class Solution {
    public String shortestPalindrome(String s) {
        if("".equals(s)) return s;
        int sum = 0, reverse = 0;
        int pos = 0;
        int power = 1;
        int mod = 19999999;
        for(int i=0;i<s.length();i++){
            int c = s.charAt(i)-'a';
            sum = (sum*26+c)%mod;
            reverse = (c*power+reverse)%mod;
            power = power*26%mod;
            if( sum==reverse && isParlindrom(s.substring(0,i+1)) ){
                pos = i;
            }
        }
        return (new StringBuilder(s.substring(pos+1)).reverse()).toString()+s;
    }
    private boolean isParlindrom(String s){
        int left = 0,right = s.length()-1;
        while(left<right){
            if(s.charAt(left) == s.charAt(right)){
                left++;right--;
            }
            else
                return false;
        }
        return true;
    }
}

647. Palindromic Substrings Medium

Given a string s, return the number of palindromic substrings in it.

A string is a palindrome when it reads the same backward as forward.

A substring is a contiguous sequence of characters within the string.

Example 1:

Input: s = "abc"
Output: 3
Explanation: Three palindromic strings: "a", "b", "c".

Example 2:

Input: s = "aaa"
Output: 6
Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa". 

Constraints:

• 1 <= s.length <= 1000
• s consists of lowercase English letters.

解法： 只需要简单的从左侧开始, 以当前字母和当前两个字母为中心向两侧判定计算palindrome

class Solution {
    int count = 0;
    public int countSubstrings(String s) {
        int len = s.length();
　　　　　//以逐个字母为中心进行计算
        for(int i=0;i<len;i++){
            valid(s,i,i);//计算奇数情况
            if(i<len-1) valid(s,i,i+1);//计算偶数个情况
        }
        return count;
    }
    public void valid(String s, int left,int right){
        int len = s.length();
        while(left>=0 && right<len){
            if(s.charAt(left)==s.charAt(right)){
                left--;right++;
                count++;
            }
            else 
                break;
        }
    }
}

 

1216. Valid Palindrome III Hard

Given a string s and an integer k, return true if s is a k-palindrome.

A string is k-palindrome if it can be transformed into a palindrome by removing at most k characters from it.

Example 1:

Input: s = "abcdeca", k = 2
Output: true
Explanation: Remove 'b' and 'e' characters.

Example 2:

Input: s = "abbababa", k = 1
Output: true

Constraints:

• 1 <= s.length <= 1000
• s consists of only lowercase English letters.
• 1 <= k <= s.length

解法：lcs，实际这个题目可以转化为longest common subsequence 问题，我们只要得到这个字符串与其反转后字符串的lcs，那么剩余的就是需要删除的个数

class Solution {
    public boolean isValidPalindrome(String s, int k) {
　　　　　//得到lcs
        int max = dfs(s,0,s.length()-1,new Integer[s.length()][s.length()]);
        return s.length()-max<=k;//长度-lcs 即为需要删除的个数
    }
    private int dfs(String s,int left,int right,Integer[][] mem){
        if(left >= s.length() || right<0) return 0;
        if(mem[left][right]!=null) return mem[left][right];
        if(s.charAt(left) == s.charAt(right)){
            mem[left][right] = dfs(s, left+1, right-1,mem)+1;
        }
        else{
            mem[left][right] = Math.max(dfs(s, left+1, right,mem),dfs(s, left, right-1,mem));
        }
        return mem[left][right];
    }
}

 

标签：palindrome,string,int,length,字符串,Input,Example,match
来源： https://www.cnblogs.com/cynrjy/p/16369454.html