lintcode594 - strStr II - hard

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.

Example

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

Rabin Karp Algorithm。用Hash方程的思想做。自己实现Hash function。

1.提前算好target的code。

2.开始读入source的每个字符，在for循环里计算所有长度为target.length()的子字符串的hashcode。

3.对比，如果code一样，要进一步对比是否substring的确一样以排除false positive.

细节：

1.第二步内比如从abc -> bcd，for循环每次都加了新的字符，d已经被加上去了，你只要轻松地排除窗口刚滑走的a*幂次即可。因为幂次固定是31*target.length()，可以提早把幂次算好。

2.对%BASE这个操作每次计算别忘了。

3.对两个各自对BASE取余过的数做减法后，可能会出来负数，这时候不好

我的实现：

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 BASE = 100000;
        int length = target.length();

        // 31^length, 31 is HASHBASE.
        int power = 1;
        for (int i = 0; i < length; i++) {
            power = (power * 31) % BASE;
        }

        int targetCode = 0;
        for (int i = 0; i < length; i++) {
            targetCode = (targetCode * 31 + target.charAt(i)) % BASE;
        }

        int sourceCode = 0;
        for (int i = 0; i < source.length(); i++) {
            sourceCode = (sourceCode * 31 + source.charAt(i)) % BASE;

            if (i >= length) {
                //注意这里。可能出现负数的地方不要直接再%，用if处理。
                sourceCode = sourceCode - (source.charAt(i - length) * power) % BASE;
                if (sourceCode < 0) {
                    sourceCode += BASE;
                }
            }

            if (i >= length - 1 && sourceCode == targetCode
                && source.substring(i - length + 1, i + 1).equals(target)) {
                return i - length + 1;
            }
        }

        return -1;
    }
}

九章实现：

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) {
        if(target == null) {
            return -1;
        }
        int m = target.length();
        if(m == 0 && source != null) {
            return 0;
        }

        if(source == null) {
            return -1;
        }
        int n = source.length();
        if(n == 0) {
            return -1;
        }

        // mod could be any big integer
        // just make sure mod * 33 wont exceed max value of int.
        int mod = Integer.MAX_VALUE / 33;
        int hash_target = 0;

        // 33 could be something else like 26 or whatever you want
        for (int i = 0; i < m; ++i) {
            hash_target = (hash_target * 33 + target.charAt(i) - ‘a‘) % mod;
            if (hash_target < 0) {
                hash_target += mod;
            }
        }

        int m33 = 1;
        for (int i = 0; i < m - 1; ++i) {
            m33 = m33 * 33 % mod;
        }

        int value = 0;
        for (int i = 0; i < n; ++i) {
            if (i >= m) {
                value = (value - m33 * (source.charAt(i - m) - ‘a‘)) % mod;
            }

            value = (value * 33 + source.charAt(i) - ‘a‘) % mod;
            if (value < 0) {
                value += mod;
            }

            if (i >= m - 1 && value == hash_target) {
                // you have to double check by directly compare the string
                if (target.equals(source.substring(i - m + 1, i + 1))) {
                    return i - m + 1;
                }
            }
        }
        return -1;
    }
}

原文地址：https://www.cnblogs.com/jasminemzy/p/9429780.html