28. Implement strStr()(KMP字符串匹配算法)

 

---

Implement strStr().

Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.

Example 1:

Input: haystack = "hello", needle = "ll"
Output: 2

 

Example 2:

Input: haystack = "aaaaa", needle = "bba"
Output: -1

class Solution {
public:

    int strStr(string haystack, string needle) {
        if (needle.size()==0) return 0;
        if (needle.size()> haystack.size()) return -1;
        
        vector<int> next(needle.size(),0);
        int j = 0;//前缀末尾，同时也是最大相同前后缀长度
        int i = 1;//后缀末尾
    
        for(int i = 1;i < needle.size();i++) {   
            while(j>0 && needle[i]!=needle[j]) {
                j = next[j-1];
            }
            if (needle[i] == needle[j]) {
                j++;
                next[i] = j;
            }
        }
            
        j = 0;
        for(int i =0; i < haystack.size();i++) {
            while(j > 0 && haystack[i]!=needle[j]) {
                j = next[j-1];
            } 
            if (haystack[i]==needle[j]) {
                j++;
            }
            if (j==needle.size()) {
                return i-needle.size()+1;
            }
        }
        return -1;

    }
};

 

https://www.bilibili.com/video/BV1M5411j7Xx/?spm_id_from=333.788.recommend_more_video.-1

暴力

class Solution {
public:
    int strStr(string haystack, string needle) {
        if (needle.size()==0) return 0;
        if (needle.size()> haystack.size()) return -1;
        for(int a = 0; a <= haystack.size()-needle.size();a++) {
            int j = 0;
            int i = a;
            while(i < haystack.size() && j < needle.size() && haystack[i]==needle[j]) {
                j++;
                i++;
            }
            if (j==needle.size()) {
                return i-needle.size();
            } else {
                j = 0;
                i = a+1;
            }
        }
        return -1;
    }
};

 

暴力解法:如果模式串匹配失败，需要回溯到模式串的起始位置

我们想可以不用回溯到起始位置，如图：
如果能确定A==B 可以直接跳到C跟d比较

问题就转化成了如何求模式串中前缀串于后缀串相等的K个

 

 

 

 

 

class Solution:
    def strStr(self, haystack, needle):
        """
        :type haystack: str
        :type needle: str
        :rtype: int
        """
        if needle =='':
            return 0
        nexts=self.caclNext(needle)
        ans = -1
        i = 0
        j = 0
        while i < len(haystack):
            if(j==-1 or haystack[i] == needle[j]):
                i += 1
                j += 1
            else:
                j = nexts[j]
            if j == len(needle):
                ans = i - len(needle)
                break
        return ans

    def caclNext(self, p):
        nexts = [0]*(len(p))
        nexts[0] = -1
        k = -1
        j = 0
        while j < len(p) - 1:
            if k == -1 or p[j] == p[k]:
                k += 1
                j += 1
                nexts[j] = k
            else:
                k = nexts[k]
        return nexts