[LeetCode-interview algorithm classic-Java implementation] [028-Implement strStr () (Implement strStr () function)],-javastrstr

[LeetCode-interview algorithm classic-Java implementation] [028-Implement strStr () (Implement strStr () function)],-javastrstr
[028-Implement strStr () (Implement the strStr () function )][LeetCode-interview algorithm classic-Java implementation] [directory indexes for all questions]Original question

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

Theme

Implement the strStr () function to determine the position of a string in another string. -1 is returned if no match exists.

Solutions

Implementation Using KMP Algorithm

Code Implementation

Algorithm Implementation class

public class Solution {    public int strStr(String haystack, String needle) {        if (haystack == null || needle == null) {            return -1;        }        if (needle.length() > haystack.length()) {            return -1;        }        if ("".equals(haystack)) {            if ("".equals(needle)) {                return 0;            } else {                return -1;            }        } else {            if ("".equals(needle)) {                return 0;            }        }        return kmpIndex(haystack, needle);    }    private int kmpIndex(String haystack, String needle) {        int i = 0;        int j = 0;        int[] next = next(needle);        while (i < haystack.length() && j < needle.length()) {            if (j == -1 || haystack.charAt(i) == needle.charAt(j)) {                ++i;                ++j;            } else {                j = next[j];            }        }        if (j == needle.length()) {            return i - j;        } else {            return -1;        }    }    private int[] next(String needle) {        int[] next = new int[needle.length()];        next[0] = -1;        int i = 0;        int j = -1;        int k = needle.length() - 1;        while (i < k) {            if (j == -1 || needle.charAt(i) == needle.charAt(j)) {                ++i;                ++j;                if (needle.charAt(i) != needle.charAt(j)) {                    next[i] = j;                } else {                    next[i] = next[j];                }            } else {                j = next[j];            }        }        return next;    }}

Evaluation Result

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Note Please refer to the following link for more information: http://blog.csdn.net/derrantcm/article/details/47052669]

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