leetcode 28. Implement strStr() 三种解法

28. Implement strStr()

 

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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 algorithmsif(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;