leetcode 28. Implement strStr() 字符串匹配KMP

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
Clarification:

What should we return when needle is an empty string? This is a great question to ask during an interview.

For the purpose of this problem, we will return 0 when needle is an empty string. This is consistent to C’s strstr() and Java’s indexOf().

思路：字符串匹配，可以很朴素，也可以用KMP

class Solution {
public:
    int strStr(string haystack, string needle){
    int m = haystack.size(), n = needle.size();
        for (int i = 0; i <= m - n; i++) {
            int j = 0;
            for (; j < n; j++) {
                if (haystack[i + j] != needle[j]) {
                    break;
                }
            }
            if (j == n) {
                return i;
            }
        }
        return -1;
    }
};

// kmp
class Solution {
public:
    int nxt[100005];
    int strStr(string haystack, string needle){
        int n=haystack.length(),m=needle.length();
        if(m<1)return 0;
        if(n<1)return -1;
        int i=0,j;
        j=nxt[0]=-1;
        while(i<m){
            while(j!=-1&&needle[i]!=needle[j])j=nxt[j];
            nxt[++i]=++j;
        }
        i=0;j=0;
        while(i<n){
            while(j!=-1&&haystack[i]!=needle[j])j=nxt[j];
            i++;j++;
            if(j>=m)return i-m;
        }
        return -1;
    }
};

// 改进的预处理
class Solution {
public:
    int nxt[100005];
    int strStr(string haystack, string needle){
        int n=haystack.length(),m=needle.length();
        if(m<1)return 0;
        if(n<1)return -1;
        int i=0,j;
        j=nxt[0]=-1;
        while(i<m){
            while(j!=-1&&needle[i]!=needle[j])j=nxt[j];
            if(needle[++i]==needle[++j])nxt[i]=nxt[j]; // improved preprocessing: nxt[i]=nxt[nxt[..nxt[i]]] until nxt[i]<0 or x[nxt[i]]!=x[i]
            else nxt[i]=j;
        }
        i=0;j=0;
        while(i<n){
            while(j!=-1&&haystack[i]!=needle[j])j=nxt[j];
            i++;j++;
            if(j>=m)return i-m;
        }
        return -1;
    }
};