题目

1.自己想的垃圾方法

class Solution {public List<Integer> findAnagrams(String s, String p) {List<Integer> res = new ArrayList<>();char[] sArray = s.toCharArray();char[] pArray = p.toCharArray();Arrays.sort(pArray);String pSorted = String.valueOf(pArray);Set<Character> pSet = new HashSet<>();for (char c : pArray) {pSet.add(c);}for (int i = 0; i <= s.length() - p.length(); ++i) {if (!pSet.contains(sArray[i])) {continue;}char[] tmpArray = new char[p.length()];for (int j = i; j - i < p.length(); ++j) {tmpArray[j - i] = sArray[j];}Arrays.sort(tmpArray);String tmp = String.valueOf(tmpArray);if (tmp.equals(pSorted)) {res.add(i);}}return res;}
}

2.滑动窗口

重点学习《算法小抄》上关于滑动窗口模板几道题的总结对比！

class Solution {public List<Integer> findAnagrams(String s, String p) {List<Integer> res = new ArrayList<>();char[] sArray = s.toCharArray();char[] pArray = p.toCharArray();int[] need = new int[256];Set<Character> set = new HashSet<>();int[] window = new int[256];for (char c : pArray) {++need[c];set.add(c);}int valid = 0,left = 0, right = 0;while (right < sArray.length) {char cur = sArray[right];++right;if (need[cur] > 0) {++window[cur];if (need[cur] == window[cur]) {++valid;}}while (right - left >= pArray.length) {if (valid == set.size()) {res.add(left);}char d = sArray[left];++left;if (need[d] > 0) {if (window[d] == need[d]) {--valid;}--window[d];}}}return res;}
}