三刷76

Author: diguage

Diff for: docs/0076-minimum-window-substring.adoc

@@ -1,20 +1,49 @@
 [#0076-minimum-window-substring]
-= 76. Minimum Window Substring
+= 76. 最小覆盖子串
 
-{leetcode}/problems/minimum-window-substring/[LeetCode - Minimum Window Substring^]
+https://leetcode.cn/problems/minimum-window-substring/[LeetCode - 76. 最小覆盖子串 ^]
 
-Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).
+给你一个字符串 `s` 、一个字符串 `t` 。返回 `s` 中涵盖 `t` 所有字符的最小子串。如果 `s` 中不存在涵盖 `t` 所有字符的子串，则返回空字符串 `""` 。
 
-.Example:
-----
-Input: S = "ADOBECODEBANC", T = "ABC"
-Output: "BANC"
-----
+*注意：*
+
+* 对于 `t` 中重复字符，我们寻找的子字符串中该字符数量必须不少于 `t` 中该字符数量。
+* 如果 `s` 中存在这样的子串，我们保证它是唯一的答案。
+
+*示例 1：*
+
+....
+输入：s = "ADOBECODEBANC", t = "ABC"
+输出："BANC"
+解释：最小覆盖子串 "BANC" 包含来自字符串 t 的 'A'、'B' 和 'C'。
+....
+
+*示例 2：*
+
+....
+输入：s = "a", t = "a"
+输出："a"
+解释：整个字符串 s 是最小覆盖子串。
+....
+
+*示例 3:*
 
-*Note:*
+....
+输入: s = "a", t = "aa"
+输出: ""
+解释: t 中两个字符 'a' 均应包含在 s 的子串中，
+因此没有符合条件的子字符串，返回空字符串。
+....
 
-* If there is no such window in S that covers all characters in T, return the empty string "".
-* If there is such window, you are guaranteed that there will always be only one unique minimum window in S.
+
+*提示：*
+
+* `m == s.length`
+* `n == t.length`
+* `1 \<= m, n \<= 10^5^`
+* `s` 和 `t` 由英文字母组成
+
+**进阶：**你能设计一个在 `O(m+n)` 时间内解决此问题的算法吗？
 
 == 思路分析
 
@@ -23,15 +52,35 @@ Output: "BANC"
 image::images/0076-01.gif[{image_attr}]
 
 [[src-0076]]
+[tabs]
+====
+一刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0076_MinimumWindowSubstring.java[tag=answer]
 ----
+--
 
+二刷::
++
+--
 [{java_src_attr}]
 ----
 include::{sourcedir}/_0076_MinimumWindowSubstring_2.java[tag=answer]
 ----
+--
+
+三刷::
++
+--
+[{java_src_attr}]
+----
+include::{sourcedir}/_0076_MinimumWindowSubstring_3.java[tag=answer]
+----
+--
+====
 
 == 参考资料
 

Diff for: logbook/202503.adoc

@@ -55,6 +55,11 @@ endif::[]
 |{doc_base_url}/0202-happy-number.adoc[题解]
 |✅ 快慢指针
 
+|{counter:codes2503}
+|{leetcode_base_url}/minimum-window-substring/[76. Minimum Window Substring^]
+|{doc_base_url}/0076-minimum-window-substring.adoc[题解]
+|⭕️ 滑动窗口，一定注意细节的处理。
+
 |===
 
 截止目前，本轮练习一共完成 {codes2503} 道题。

Diff for: src/main/java/com/diguage/algo/leetcode/_0076_MinimumWindowSubstring_2.java

@@ -16,6 +16,9 @@ public class _0076_MinimumWindowSubstring_2 {
 
   /**
    * 自己实现失败后，参考社区答案修改的
+   *
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2024-07-02 17:03:23
    */
   public String minWindow(String s, String t) {
     if (s == null || s.isEmpty() || t == null || t.isEmpty() || s.length() < t.length()) {

Diff for: src/main/java/com/diguage/algo/leetcode/_0076_MinimumWindowSubstring_3.java

@@ -0,0 +1,48 @@
+package com.diguage.algo.leetcode;
+
+import java.util.HashMap;
+import java.util.Map;
+import java.util.Objects;
+
+public class _0076_MinimumWindowSubstring_3 {
+  // tag::answer[]
+  /**
+   * @author D瓜哥 · https://www.diguage.com
+   * @since 2025-03-21 11:08:35
+   */
+  public String minWindow(String s, String t) {
+    if (s == null || t == null || s.isEmpty() || t.isEmpty() || s.length() < t.length()) {
+      return "";
+    }
+    Map<Character, Integer> target = new HashMap<>();
+    for (char c : t.toCharArray()) {
+      target.put(c, target.getOrDefault(c, 0) + 1);
+    }
+    int left = 0, right = 0;
+    int valid = 0, startIdx = 0, minLength = Integer.MAX_VALUE;
+    Map<Character, Integer> windows = new HashMap<>();
+    while (right < s.length()) {
+      char rc = s.charAt(right);
+      right++;
+      windows.put(rc, windows.getOrDefault(rc, 0) + 1);
+      if (target.containsKey(rc)
+        && Objects.equals(target.get(rc), windows.get(rc))) {
+        valid++;
+      }
+      while (valid == target.size()) {
+        if (right - left < minLength) {
+          minLength = right - left;
+          startIdx = left;
+        }
+        char lc = s.charAt(left);
+        windows.put(lc, windows.getOrDefault(lc, 0) - 1);
+        if (target.containsKey(lc) && windows.get(lc) < target.get(lc)) {
+          valid--;
+        }
+        left++;
+      }
+    }
+    return minLength == Integer.MAX_VALUE ? "" : s.substring(startIdx, startIdx + minLength);
+  }
+  // end::answer[]
+}