LintCode 647 Substring Anagrams

最新推荐文章于 2020-09-19 11:55:19 发布

IncludeLayout

最新推荐文章于 2020-09-19 11:55:19 发布

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本文介绍了一种高效算法来解决寻找一个字符串中所有由另一字符串颠倒字母顺序构成的子串的问题。该算法利用哈希映射来跟踪目标字符串的字符频率，并通过滑动窗口方式遍历主字符串，以此找到所有符合条件的起始索引。

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Given a string s and a non-empty string p, find all the start indices of p’s anagrams in s.

Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 40,000.

The order of output does not matter.

样例
Given

s = "cbaebabacd" p = "abc"

return [0, 6]

The substring with start index = 0 is “cba”, which is an anagram of “abc”.
The substring with start index = 6 is “bac”, which is an anagram of “abc”.

题目的意思是给出字符串s和p，在s字符串中找出所有由p字符串颠倒字母顺序而构成的字符串，输出结果字符串的开始索引。
解一：

/**
     * @param s a string
     * @param p a non-empty string
     * @return a list of index
     */
    public List<Integer> findAnagrams(String s, String p) {
        ArrayList<Integer> result = new ArrayList<>();
        if (s == null || p == null) {
            return result;
        }
        int left = 0, right = 0, count = p.length();
        int map[] = new int[256];
        char[] sc = s.toCharArray();
        for (char c : p.toCharArray()) {
            map[c]++;
        }
        while (right < s.length()) {
            if (map[sc[right++]]-- >= 1) {
                count--;
            }
            if (count == 0) {
                result.add(left);
            }
            if (right - left == p.length() && map[sc[left++]]++ >= 0) {
                count++;
            }
        }
        return result;
    }