力扣labuladong一刷day21天滑动哈希算法共2题

力扣labuladong一刷day21天滑动哈希算法共2题

一、187. 重复的DNA序列

题目链接：https://leetcode.cn/problems/repeated-dna-sequences/description/
思路：字符串序列里找重复出现的子串就是子串匹配问题，正常的思路是截取所有长为L的子串，看set里是否重复，可是截取的过程是O(n)每个都截取就成了O(x*n)。采用另外一种思路，也是滑动窗口，只不过把滑动窗口中的字符串映射成数字，这种映射的方式就是right右移尾部加数，left右移头部减数，复杂度是O(1)。这样只需要比较set中的映射值即可。

class Solution {
    public List<String> findRepeatedDnaSequences(String s) {
        Set<String> list = new HashSet<>();
        Set<Integer> set = new HashSet<>();
        int[] nums = new int[s.length()];
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (c == 'A') {
                nums[i] = 0;
            } else if (c == 'C') {
                nums[i] = 1;
            } else if (c == 'G') {
                nums[i] = 2;
            }else {
                nums[i] = 3;
            }
        }
        int hashcode  = 0, left = 0, right = 0, r = 4;
        int rl = (int) Math.pow(r,9);
        while (right < nums.length) {
            hashcode = hashcode * r + nums[right];
            right++;
            if (right - left == 10) {
                if (set.contains(hashcode)) {
                    list.add(s.substring(left, right));
                }else {
                    set.add(hashcode);
                }
                hashcode = hashcode - nums[left] * rl;
                left++;
            }
        }
        return new ArrayList<>(list);
    }
}

二、28. 找出字符串中第一个匹配项的下标

题目链接：https://leetcode.cn/problems/find-the-index-of-the-first-occurrence-in-a-string/
思路：滑动窗口使用needmap和windowmap要求需要匹配的是子串的排列，如果不是排列是无法区分ab与ba的。本题是完全匹配就只能用滑动哈希算法了。

选择一个较大的素数Q，这样会减少哈希碰撞。

class Solution {
   public int strStr(String haystack, String needle) {
        long Q = 1658598167, L = needle.length(), R = 256;
        long hashCode = 0, patCode = 0;
        long RL = 1;
        for (int i = 0; i < L - 1; i++) {
            RL = (RL*R) % Q;
        }
        for (int i = 0; i < L; i++) {
            patCode = (patCode * R+ needle.charAt(i)) % Q;
        }
        int left = 0, right = 0;
        while (right < haystack.length()) {
            hashCode = ((hashCode * R) % Q + haystack.charAt(right)) % Q;
            right++;
            if (right - left == L) {
                if (hashCode == patCode) {
                    if (needle.equals(haystack.substring(left, right))) return left;
                }
                hashCode = (hashCode - (haystack.charAt(left)*RL) % Q + Q) % Q;
                left++;
            }
        }
        return -1;
    }
}