LeetCode 28. Implement strStr()

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mb5fca0b6a49e47 2015-09-20 05:19:00

文章标签 LeetCode String KMP java i++ 文章分类 Java 后端开发

原题链接在这里：https://leetcode.com/problems/implement-strstr/

题目:

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().

题解：

Brute force的算法是从haystack头开始走，每次取后面长度与needle长度相等的substring, 看这个substring与needle是否相同即可.

If needle is empty, return 0.

Note: s.substring(i), i could be s.length(), and return "". Thus here, it is allowed to use i <= m - n.

Time Complexity: O(m*n), m = haystack.length(), n = needle.length().

Space: O(1).　　　

AC Java:

 1 class Solution {
 2     public int strStr(String haystack, String needle) {
 3         if(haystack == null || needle == null){
 4             return -1;
 5         }
 6         
 7         if(needle.length() == 0){
 8             return 0;
 9         }
10         
11         int m = haystack.length();
12         int n = needle.length();
13         
14         for(int i = 0; i <= m - n; i++){
15             if(haystack.substring(i, i + n).equals(needle)){
16                 return i;
17             }
18         }
19         
20         return -1;
21     }
22 }

本题还可以采用KMP算法，这篇帖子说的很详细。

注意在求next数组时，while loop的条件是q<len-1, 因为下面还用到q++后的next[q]. next array的含义是到当前char的前一位的最大公共前缀后缀.

Time Complexity: O(m+n). m = haystack.length(), n = needle.length(). Space: O(n)

AC Java:

 1 public class Solution {
 2     public int strStr(String haystack, String needle) {
 3         if(haystack == null || needle == null || haystack.length() < needle.length()){
 4             return -1;
 5         }
 6         
 7         if(needle.length() == 0){
 8             return 0;
 9         }
10         
11         int i = 0;
12         int j = 0;
13         int len1 = haystack.length();
14         int len2 = needle.length();
15         int [] next = new int[len2];
16         getNext(needle, next);
17         while(i<len1 && j<len2){
18             if(j == -1 || haystack.charAt(i) == needle.charAt(j)){
19                 i++;
20                 j++;
21             }else{
22                 j = next[j];
23             }
24         }
25         
26         if(j == len2){
27             return i-j;
28         }else{
29             return -1;
30         }
31     }
32     
33     private void getNext(String s, int [] next){
34         int len = s.length();
35         next[0] = -1;
36         int p = -1;
37         int q = 0;
38         while(q < len-1){
39             //s.charAt(p)表示前缀. s.charAt(q)表示后缀
40             if(p == -1 || s.charAt(p) == s.charAt(q)){
41                 p++;
42                 q++;
43                 next[q] = p;
44             }else{
45                 p = next[p];
46             }
47         }
48     }
49 }