【leetcode】【easy】28. Implement strStr()

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本文深入解析LeetCode上的经典字符串查找题目——实现strStr()函数，探讨了两种主流解决策略：暴力匹配与KMP算法，详细阐述了各自的实现原理与时间复杂度。

28. Implement strStr()

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().

题目链接：https://leetcode-cn.com/problems/implement-strstr/

思路

法一：暴力

对主串中每个字符作为起始，匹配是否和模式串相等。

时间复杂度O(n*m)

法二：kmp

kmp方法的经典题目。

kmp忘了的话可以从这里复习下：https://www.bilibili.com/video/av93881213?p=33

时间复杂度O(n+m)

class Solution {
public:
    int strStr(string haystack, string needle) {
        int len1 = haystack.size(), len2 = needle.size();
        if(len2==0) return 0;
        if(len1==0 ||len1<len2) return -1;
        int next[len2+1];
        getNext(needle, next);
        int i=0,j=0;
        while(i<len1 && j<len2){
            if(j<0 || haystack[i]==needle[j]){
                ++i;
                ++j;
            }else{
                j = next[j];
            }
        }
        if(j>=len2) return i-len2;
        else return -1;
    }
    void getNext(string s, int next[]){
        next[0] = -1;
        int i = -1, j = 0;
        while(j<s.size()){
            if(i<0 || s[i]==s[j]){
                next[++j] = ++i;
            }else{
                i = next[i];
            }
        }
    }
};