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第二十课字符串

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字符串

基础知识

x = ‘abbc’ x = “abbc”

String x = “abbc”;
C++:

string x(“abbc”);

遍历字符串

for ch in “abbc”:   print(ch)
String x = “abbc”; for (int i = 0; i < x.size(); ++i) {   char ch = x.charAt(i); } for ch in x.toCharArray() {   System.out.println(ch); }
string x(“abbc”); for (int i = 0; i < s1.length(); i++) 
{   
    cout << x[i]; 
}

字符串比较

Java:

String x = “abb”;
String y = “abb”;

x == y —-> false

x.equals(y) —-> true
x.equalsIgnoreCase(y) —-> true

字符串相关算法

基础问题

  1. https://leetcode-cn.com/problems/to-lower-case/

  2. https://leetcode-cn.com/problems/length-of-last-word/

  3. https://leetcode-cn.com/problems/jewels-and-stones/

  4. https://leetcode-cn.com/problems/first-unique-character-ina-string/

  5. https://leetcode-cn.com/problems/string-to-integer-atoi/

Atoi

java版本

public int myAtoi(String str) {
    int index = 0, sign = 1, total = 0;
    //1. Empty string
    if(str.length() == 0) return 0;

    //2. Remove Spaces
    while(str.charAt(index) == ' ' && index < str.length())
        index ++;

    //3. Handle signs
    if(str.charAt(index) == '+' || str.charAt(index) == '-'){
        sign = str.charAt(index) == '+' ? 1 : -1;
        index ++;
    }
    
    //4. Convert number and avoid overflow
    while(index < str.length()){
        int digit = str.charAt(index) - '0';
        if(digit < 0 || digit > 9) break;

        //check if total will be overflow after 10 times and add digit
        if(Integer.MAX_VALUE/10 < total ||            
        	Integer.MAX_VALUE/10 == total && Integer.MAX_VALUE %10 < digit)
            return sign == 1 ? Integer.MAX_VALUE : Integer.MIN_VALUE;

        total = 10 * total + digit;
        index ++;
    }
    return total * sign;
}

Python版本

class Solution(object):

    def myAtoi(self, s):

        if len(s) == 0 : return 0
        ls = list(s.strip())
        
        sign = -1 if ls[0] == '-' else 1

        if ls[0] in ['-','+'] : del ls[0]

        ret, i = 0, 0

        while i < len(ls) and ls[i].isdigit() :
            ret = ret*10 + ord(ls[i]) - ord('0')
            i += 1

        return max(-2**31, min(sign * ret,2**31-1))

字符串操作问题

  1. https://leetcode-cn.com/problems/longest-common-prefix/ description/

  2. https://leetcode-cn.com/problems/reverse-string https://leetcode-cn.com/problems/reverse-string-ii/

  3. https://leetcode-cn.com/problems/reverse-words-in-a-string/ https://leetcode-cn.com/problems/reverse-words-in-a-stringiii/

  4. https://leetcode-cn.com/problems/reverse-only-letters/

Anagram异位词问题

  1. https://leetcode-cn.com/problems/valid-anagram/

  2. https://leetcode-cn.com/problems/group-anagrams/

  3. https://leetcode-cn.com/problems/find-all-anagrams-in-astring/
    Anagram

Palindrome 回文串问题

  1. https://leetcode-cn.com/problems/valid-palindrome/

  2. https://leetcode-cn.com/problems/valid-palindrome-ii/

  3. https://leetcode-cn.com/problems/longest-palindromicsubstring/
    Palindrome

高级字符串算法

最长子串、子序列

  1. Longest common sequence(最长子序列) https://leetcode-cn.com/problems/longest-common-subsequence/

dp[i][j] = dp[i-1][j-1] + 1 (if s1[i-1] == s2[j-1]) else dp[i][j] = max(dp[i-1][j], dp[i][j-1])

  1. Longest common substring (最长子串) dp[i][j] = dp[i-1][j-1] + 1 (if s1[i-1] == s2[j-1]) else dp[i][j] = 0

  2. Edit distance(编辑距离) https://leetcode-cn.com/problems/edit-distance/

最长公共子序列

  1. 暴力 O(n^3)

2.中间向两边扩张法 O(n^2)

3.动态规划

首先定义 P(i, j):

true s[i, j] 是回文串

false s[i, j] 不是a回文串

接下来

P(i, j) = (P(i+1, j-1) && S[i] == S[j])

字符串+递归 or DP

https://leetcode-cn.com/problems/regular-expressionmatching/

重点: https://leetcode-cn.com/problems/regular-expressionmatching/solution/ji-yu-guan-fang-ti-jie-gen-xiang-xi-de-jiangjie-b/

https://leetcode-cn.com/problems/wildcard-matching/

字符串匹配算法

  1. 暴力法(brute force) - O(mn)

2.Rabin-Karp 算法

3.KMP 算法

暴力法

public final static int D = 256;
public final static int Q = 9997;

static int RabinKarpSerach(String txt, String pat) {
    int M = pat.length();
    int N = txt.length();
    int i, j;
    int patHash = 0, txtHash = 0;

    for (i = 0; i < M; i++) {
        patHash = (D * patHash + pat.charAt(i)) % Q;
        txtHash = (D * txtHash + txt.charAt(i)) % Q;
    }

    int highestPow = 1;  // pow(256, M-1)
    for (i = 0; i < M - 1; i++) 
        highestPow = (highestPow * D) % Q;

    for (i = 0; i <= N - M; i++) { // 枚举起点
        if (patHash == txtHash) {
            for (j = 0; j < M; j++) {
                if (txt.charAt(i + j) != pat.charAt(j))
                    break;
            }
            if (j == M)
                return i;
        }
        if (i < N - M) {
            txtHash = (D * (txtHash - txt.charAt(i) * highestPow) + txt.charAt(i + M)) % Q;
            if (txtHash < 0)
                txtHash += Q;
        }
    }

    return -1;
}

Rabin-Karp 算法

在朴素算法中,我们需要挨个比较所有字符,才知道目标字符串中是否包含 子串。那么, 是否有别的方法可以用来判断目标字符串是否包含子串呢?

答案是肯定的,确实存在一种更快的方法。为了避免挨个字符对目标字符串 和子串进行比较, 我们可以尝试一次性判断两者是否相等。因此,我们需 要一个好的哈希函数(hash function)。 通过哈希函数,我们可以算出子 串的哈希值,然后将它和目标字符串中的子串的哈希值进行比较。 这个新 方法在速度上比暴力法有显著提升。

Rabin-Karp 算法的思想:

  1. 假设子串的长度为 M (pat),目标字符串的长度为 N (txt)

  2. 计算子串的 hash 值 hash_pat

  3. 计算目标字符串txt中每个长度为 M 的子串的 hash 值(共需要计算 N-M+1 次)

  4. 比较 hash 值:如果 hash 值不同,字符串必然不匹配; 如果 hash 值相同, 还需要使用朴素算法再次判断

KMP算法

KMP算法(Knuth-Morris-Pratt)的思想就是,当子串与目标字符串不匹配时, 其实你已经知道了前面已经匹配成功那 一部分的字符(包括子串与目标字符 串)。以阮一峰的文章为例,当空格与 D 不匹配时,你其实 知道前面六个字符是 “ABCDAB”。KMP 算法的想法是,设法利用这个已知信息,不要把“搜索位 置” 移回已经比较过的位置,继续把它向后移,这样就提高了效率。

https://www.bilibili.com/video/ av11866460?from=search&seid=17425875345653862171

http://www.ruanyifeng.com/blog/2013/05/ Knuth%E2%80%93Morris%E2%80%93Pratt_algorithm.html

标签:cn,int,problems,https,字符串,第二十,com,leetcode
来源: https://www.cnblogs.com/liugangjiayou/p/12638995.html