LeetCode-28. Implement strStr

LeetCode-28. Implement strStr()https://leetcode.com/problems/implement-strstr/

Implement strStr().

Given two strings needle and haystack, 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().

Example 1:

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

Example 2:

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

Constraints:

• 1 <= haystack.length, needle.length <= 10^4
• haystack and needle consist of only lowercase English characters.

【C++】

class Solution {
public:
    int strStr(string haystack, string needle) {
        int k = -1, n = haystack.length(), p = needle.length();
        if (p == 0) {return 0;}
        vector next(p, -1); // -1表示不存在相同的最大前缀和后缀
        calNext(needle, next); // 计算next数组
        for (int i = 0; i < n; ++i) {
            // 有部分匹配，往前回溯
            while (k > -1 && needle[k+1] != haystack[i]) {k = next[k];}
            if (needle[k+1] == haystack[i]) {++k;}
            // 说明k移动到needle的最末端，返回相应的位置
            if (k == p-1) {return i-p+1;}
        }
        return -1;
    }

    // 辅函数- 计算next数组
    void calNext(const string &needle, vector &next) {
        for (int j = 1, p = -1; j < needle.length(); ++j) {
            // 如果下一位不同，往前回溯
            while (p > -1 && needle[p+1] != needle[j]) {p = next[p];}
            // 如果下一位相同，更新相同的最大前缀和最大后缀长
            if (needle[p+1] == needle[j]) {++p;}
            next[j] = p;
        }
    }
};

【Java】

class Solution {
    public int strStr(String haystack, String needle) {
        int k = -1, n = haystack.length(), p = needle.length();
        if (p == 0) {return 0;}
        int[] next = new int[p];
        Arrays.fill(next, -1); // -1表示不存在相同的最大前缀和后缀
        calNext(needle, next); // 计算next数组
        for (int i = 0; i < n; ++i) {
            // 有部分匹配，往前回溯
            while (k > -1 && needle.charAt(k+1) != haystack.charAt(i)) {k = next[k];}
            if (needle.charAt(k+1) == haystack.charAt(i)) {++k;}
            // 说明k移动到needle的最末端，返回相应的位置
            if (k == p-1) {return i-p+1;}
        }
        return -1;
    }

    // 辅函数- 计算next数组
    void calNext(String needle, int[] next) {
        for (int j = 1, p = -1; j < needle.length(); ++j) {
            // 如果下一位不同，往前回溯
            while (p > -1 && needle.charAt(p+1) != needle.charAt(j)) {p = next[p];}
            // 如果下一位相同，更新相同的最大前缀和最大后缀长
            if (needle.charAt(p+1) == needle.charAt(j)) {++p;}
            next[j] = p;
        }
    }
     
}

参考文献

【1】【算法图文动画详解系列】KMP 字串匹配搜索算法_东海陈光剑的博客-CSDN博客