Leetcode No.28 Implement strStr()字符串匹配(c++实现）

1. 题目

1.1 英文题目

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().

1.2 中文题目

给定一个 haystack 字符串和一个 needle 字符串，在 haystack 字符串中找出 needle 字符串出现的第一个位置 (从0开始)。如果不存在，则返回 -1。

1.3输入输出

输入	输出
haystack = "hello", needle = "ll"	2
haystack = "aaaaa", needle = "bba"	-1
haystack = "", needle = ""	0

1.4 约束条件

0 <= haystack.length, needle.length <= 5 * 104
haystack and needle consist of only lower-case English characters.

2. 分析

2.1 Brute Force（暴力搜索）算法

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

2.2 KMP算法

算法解释可参考：
KMP字符串匹配算法2
从头到尾彻底理解KMP
代码如下：

class Solution {
public:
    int strStr(string haystack, string needle) {
        int m = haystack.size(), n = needle.size();
        if (!n) {
            return 0;
        }
        vector<int> lps = kmpProcess(needle);
        for (int i = 0, j = 0; i < m;) {
            if (haystack[i] == needle[j]) {
                i++, j++;
            }
            if (j == n) {
                return i - j;
            }
            if (i < m && haystack[i] != needle[j]) {
                j ? j = lps[j - 1] : i++;
            }
        }
        return -1;
    }
private:
    vector<int> kmpProcess(string needle) {
        int n = needle.size();
        vector<int> lps(n, 0);
        for (int i = 1, len = 0; i < n;) {
            if (needle[i] == needle[len]) {
                lps[i++] = ++len;
            }
            else if (len) {
                len = lps[len - 1];
            }
            else {
                lps[i++] = 0;
            }
        }
        return lps;
    }
};