leetcode 28. Implement strStr() 三种解法

28. Implement strStr()

 

Question
Editorial Solution
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Total Accepted: 119996
Total Submissions: 467056
Difficulty: Easy

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

作弊解法：

public class Solution {
public int strStr(String haystack, String needle) {
return haystack.indexOf(needle);
}
}

子串匹配解法：

public class Solution {
public int strStr(String haystack, String needle) {
int l1 = haystack.length(), l2 = needle.length();
if (l1 < l2) {
return -1;
} else if (l2 == 0) {
return 0;
}
int threshold = l1 - l2;
for (int i = 0; i <= threshold; ++i) {
if (haystack.substring(i,i+l2).equals(needle)) {
return i;
}
}
return -1;
}
}

KMP解法：（真的解法）

public String strStr(String haystack, String needle) {
//KMP algorithms
if(needle.equals("")) return haystack;
if(haystack.equals("")) return null;
char[] arr = needle.toCharArray();
int[] next = makeNext(arr);

for(int i = 0, j = 0, end = haystack.length(); i < end;){
if(j == -1 || haystack.charAt(i) == arr[j]){
j++;
i++;
if(j == arr.length) return haystack.substring(i - arr.length);
}
if(i < end && haystack.charAt(i) != arr[j]) j = next[j];
}
return null;
}

private int[] makeNext(char[] arr){
int len = arr.length;
int[] next = new int[len];

next[0] = -1;
for(int i = 0, j = -1; i + 1 < len;){
if(j == -1 || arr[i] == arr[j]){
next[i+1] = j+1;
if(arr[i+1] == arr[j+1]) next[i+1] = next[j+1];
i++;
j++;
}
if(arr[i] != arr[j]) j = next[j];
}

return next;