力扣labuladong一刷day9滑动窗口共4题

力扣labuladong一刷day9滑动窗口共4题

一、76. 最小覆盖子串

题目链接：https://leetcode.cn/problems/minimum-window-substring/
思路：典型的滑动窗口题目，使用一个map记录所必须的字符个数，使用另外一个map去记录滑动窗口内部的need字符，一旦need所需的个数都满足以后，就开始缩小滑动窗口，在缩小滑动窗口的过程中不断记录最小的窗口长度以及窗口的起始点，并且在map不满足need时结束缩小窗口，继续扩大窗口。

class Solution {public String minWindow(String s, String t) {Map<Character, Integer> need = new HashMap<>();Map<Character, Integer> window = new HashMap<>();int left = 0, right = 0, valid = 0;int start = 0, max = Integer.MAX_VALUE;for (char c : t.toCharArray()) {need.put(c, need.getOrDefault(c, 0)+1);}while (right < s.length()) {char c = s.charAt(right);right++;if (need.containsKey(c)) {window.put(c, window.getOrDefault(c, 0)+1);if (window.get(c).equals(need.get(c))) {valid++;}}while (valid == need.size()) {if (right - left < max) {start = left;max = right-start;}char cl = s.charAt(left);left++;if (need.containsKey(cl)) {if (window.get(cl).equals(need.get(cl))) valid--;window.put(cl, window.get(cl)-1);}}}return max == Integer.MAX_VALUE ? "" : s.substring(start, start+max);}
}

二、567. 字符串的排列

题目链接：https://leetcode.cn/problems/permutation-in-string/
思路：本题要求s1是s2子串的排列，那就是要求s1与s2的子串长度要相等，那就是我们只需要控制滑动窗口的长度等于子串长度即可，长度相等时只要s1中的字符都出现了即可返回，然后就是正常缩小窗口，再扩大窗口。

class Solution {public boolean checkInclusion(String s1, String s2) {Map<Character, Integer> need = new HashMap<>();Map<Character, Integer> window = new HashMap<>();int left = 0, right = 0, valid = 0;for (char c : s1.toCharArray()) {need.put(c, need.getOrDefault(c, 0) + 1);}while (right < s2.length()) {char c = s2.charAt(right);right++;if (need.containsKey(c)) {window.put(c, window.getOrDefault(c, 0)+1);if (window.get(c).equals(need.get(c))) valid++;}while (right - left == s1.length()) {if (valid == need.size()) {return true;}char cl = s2.charAt(left);left++;if (need.containsKey(cl)) {if (window.get(cl).equals(need.get(cl))) valid--;window.put(cl, window.get(cl)-1);}}}return false;}
}

三、438. 找到字符串中所有字母异位词

题目链接：https://leetcode.cn/problems/find-all-anagrams-in-a-string/
思路：和上一题基本差不多，也是要求p与s的子串长度相等，我们只需要控制窗口等于p的长度即可，然后在其中判断。

class Solution {public List<Integer> findAnagrams(String s, String p) {List<Integer> list = new ArrayList<>();Map<Character, Integer> need = new HashMap<>();Map<Character, Integer> window = new HashMap<>();int left = 0, right = 0, valid = 0;for (char c : p.toCharArray()) {need.put(c, need.getOrDefault(c, 0)+1);}while (right < s.length()) {char cr = s.charAt(right);right++;if (need.containsKey(cr)) {window.put(cr, window.getOrDefault(cr, 0)+1);if (window.get(cr).equals(need.get(cr))) valid++;}if (right - left == p.length()) {if (valid == need.size()) {list.add(left);}char cl = s.charAt(left);left++;if (need.containsKey(cl)) {if (window.get(cl).equals(need.get(cl))) valid--;window.put(cl, window.get(cl)-1);}}}return list;}
}

四、3. 无重复字符的最长子串

题目链接：https://leetcode.cn/problems/longest-substring-without-repeating-characters/
思路：求无重复字符串的最长子串，只需要用map收集字符即可，只要当前字符个数大于1即可开始缩小滑动窗口，直到当前字符的个数不再大于1.
最大长度max的记录放在最后。

class Solution {public int lengthOfLongestSubstring(String s) {Map<Character, Integer> map = new HashMap<>();int left = 0, right = 0, max = 0;while (right < s.length()) {char c = s.charAt(right);right++;map.put(c, map.getOrDefault(c, 0)+1);while (map.get(c) > 1) {char cl = s.charAt(left);left++;map.put(cl, map.get(cl)-1);}max = Math.max(max, right - left);}return max;}
}