438. Find All Alphabetic Words in a String
Given two strings s and p, find all anagram substrings of p in s, and return the starting indices of these substrings. The order in which answers are output is not considered.
An anagram refers to a character string formed by rearranging the same letter (including the same character string).
Example 1:
输入: s = "cbaebabacd", p = "abc"
输出: [0,6]
解释:
起始索引等于 0 的子串是 "cba", 它是 "abc" 的异位词。
起始索引等于 6 的子串是 "bac", 它是 "abc" 的异位词。
Example 2:
输入: s = "abab", p = "ab"
输出: [0,1,2]
解释:
起始索引等于 0 的子串是 "ab", 它是 "ab" 的异位词。
起始索引等于 1 的子串是 "ba", 它是 "ab" 的异位词。
起始索引等于 2 的子串是 "ab", 它是 "ab" 的异位词。
hint:
1 <= s.length, p.length <= 3 * 104
s 和 p 仅包含小写字母
answer:
class Solution {
public List<Integer> findAnagrams(String s, String p) {
List<Integer> result = new ArrayList<>();
// 使用双指针+位图
int n = s.length();
int m = p.length();
//
if (m > n) {
return new ArrayList<>();
}
// p 在位图中的位置
int[] needs = new int[26];
for (int i=0;i<m;i++) {
needs[p.charAt(i) - 'a']++;
}
// 使用窗口滑动 s
int start = 0;
int end = 0;
int[] matched = new int[26]; // 窗口
// 先比较前m个字符
while (end < m) {
matched[s.charAt(end)-'a']++;
end++;
}
if (same(needs, matched)) {
result.add(start);
}
while (start<n && end<n) {
// 滑动窗口
matched[s.charAt(start)-'a']--;
matched[s.charAt(end)-'a']++;
start++;
end++;
if (same(needs, matched)) {
result.add(start);
}
}
return result;
}
public boolean same(int[] needs, int[] matched) {
for (int i=0;i<needs.length;i++) {
if (needs[i] != matched[i]) {
return false;
}
}
return true;
}
}