#Question
Implement strStr().Return the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack.
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().

Example1:

Input: haystack = “hello”, needle = “ll”
Output: 2

Example2:

Input: haystack = “aaaaa”, needle = “bba”
Output: -1

Example3:

Input: haystack = “”, needle = “”
Output: 0

Answer

##思路一:
两个字符长度相减+1得到的长度就是需要比较的长度，当needle长度为0时，返回0。当haystack与needle不匹配时，返回-1。当haystack与needle匹配时，返回i的值。时间复杂度为O($n^{2}$),空间复杂度为O(n),具体代码如下：

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

##思路二:
KMP(字符串查找算法),lps是needle的索引。这里使用的kmp算法，把needle作为kmp前缀,来进行字符比对。如果字符相同，那么len增加。如果不相等，就要看len的值，如果len的值为0那么就将lps[i]的值设置为0。如果len不为零，那么就把lps[len-1]的值赋给len，这一步是为了防止对应字符中间出现了不相等的，将字符串右移。时间复杂度为O(n);

class Solution {
public:
    int strStr(string haystack, string needle) {
        if (needle.empty()) return 0;
        
        string kmp = needle + "#" + haystack;
        int m = kmp.size(),n = needle.size();
        vector<int> lps(m,0);
        
        for (int len = 0,i = 1; i < m;) {
            if (kmp[len] == kmp[i]) {
                lps[i] = ++len;
                if (len == n) 
                    return i - 2 * n;
                ++i;
            }
            else if (len == 0) 
                 lps[i++] = 0;
            else 
                len = lps[len - 1];
        }
        return -1;
    }
};

Rabbet

实现strStr()

Answer