strStr II

Implement strStr function in O(n + m) time.

strStr return the first index of the target string in a source string. The length of the target string is m and the length of the source string is n.
If target does not exist in source, just return -1.

Have you met this question in a real interview? 

Yes

Example

Given source = abcdef, target = bcd, return 1.

每进行一次hash运算的时候，都需要进行取模处理；

java

public class Solution {
    /*
     * @param source: A source string
     * @param target: A target string
     * @return: An integer as index
     */
    public int strStr2(String source, String target) {
        // write your code here
        if (source == null || target == null) {
            return -1;
        }
        if (target.length() == 0) {
            return 0;
        }
        int sourceLen = source.length();
        int targetLen = target.length();
        int sourceVal = 0;
        int targetVal = 0;
        int BASE = 10000;
        int magic = 33;
        int val = 1;
        for (int i = 0; i < targetLen; i++) {
            targetVal = targetVal * magic + target.charAt(i) - 'a';
            targetVal %= BASE;
        }
        for (int i = 0; i < targetLen - 1; i++) {
            val *= magic;
            val %= BASE;
        }
        for (int i = 0; i < sourceLen; i++) {
            if (i >= targetLen) {
                sourceVal -= (source.charAt(i - targetLen) - 'a') * val;
                sourceVal %= BASE;
                if (sourceVal < 0) {
                    sourceVal += BASE;
                }
            }
            sourceVal = sourceVal * magic + source.charAt(i) - 'a';
            if (i >= targetLen - 1 && sourceVal == targetVal) {
                if (target.equals(source.substring(i - targetLen + 1, i + 1))) {
                    return i - targetLen + 1;
                }
            }
        }
        return -1;
    }
}