[LeetCode] 438. Find All Anagrams in a String

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

Strings consist of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.

The order of output does not matter.

Example 1:

Input:
s: "cbaebabacd" p: "abc"

Output:
[0, 6]

Explanation:
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".

Example 2:

Input:
s: "abab" p: "ab"

Output:
[0, 1, 2]

Explanation:
The substring with start index = 0 is "ab", which is an anagram of "ab".
The substring with start index = 1 is "ba", which is an anagram of "ab".
The substring with start index = 2 is "ab", which is an anagram of "ab".

找到字符串中所有字母易位词。给定一个字符串 s 和一个非空字符串 p，找到 s 中所有是 p 的字母异位词的子串，返回这些子串的起始索引。思路是滑动窗口（sliding window）。这个题也可以用之前的模板做，参见76，159，340。先遍历p，得到p中每个字母和他们对应的出现次数。再去遍历s，用start和end指针，end在前，当某个字母的出现次数被减少到0的时候，counter才能--。当counter == 0的时候，判断end - begin是否等于p的长度，若是，才能将此时begin的坐标加入结果集。

时间O(n)

空间O(1)

Java实现

class Solution {
    public List<Integer> findAnagrams(String s, String p) {
        List<Integer> res = new ArrayList<>();
        // corner case
        if (s.length() < p.length()) {
            return res;
        }

        // normal case
        int[] map = new int[128];
        for (char c : p.toCharArray()) {
            map[c]++;
        }
        int counter = p.length();
        int start = 0;
        int end = 0;
        while (end < s.length()) {
            char c1 = s.charAt(end);
            if (map[c1] > 0) {
                counter--;
            }
            map[c1]--;
            end++;
            while (counter == 0) {
                char c2 = s.charAt(start);
                map[c2]++;
                if (map[c2] > 0) {
                    counter++;
                }
                if (end - start == p.length()) {
                    res.add(start);
                }
                start++;
            }
        }
        return res;
    }
}

JavaScript实现

/**
 * @param {string} s
 * @param {string} p
 * @return {number[]}
 */
var findAnagrams = function(s, p) {
    let res = [];
    if (p.length > s.length) {
        return res;
    }

    let map = new Map();
    for (let c of p) {
        map.set(c, (map.get(c) | 0) + 1);
    }

    let begin = 0,
        end = 0,
        counter = map.size;
    while (end < s.length) {
        let c = s.charAt(end);
        if (map.has(c)) {
            map.set(c, map.get(c) - 1);
            if (map.get(c) === 0) {
                counter--;
            }
        }
        end++;
        while (counter === 0) {
            let d = s.charAt(begin);
            if (map.has(d)) {
                map.set(d, map.get(d) + 1);
                if (map.get(d) > 0) {
                    counter++;
                }
            }
            if (end - begin == p.length) {
                res.push(begin);
            }
            begin++;
        }
    }
    return res;
};

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