From 047cd5b172ead356dd894fdb16c84ad799f15dd8 Mon Sep 17 00:00:00 2001
From: huangge1199 <huangge1199@hotmail.com>
Date: Wed, 1 Sep 2021 16:39:01 +0800
Subject: [PATCH] =?UTF-8?q?424:=E6=9B=BF=E6=8D=A2=E5=90=8E=E7=9A=84?=
 =?UTF-8?q?=E6=9C=80=E9=95=BF=E9=87=8D=E5=A4=8D=E5=AD=97=E7=AC=A6?=
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---
 .../LongestRepeatingCharacterReplacement.java | 56 +++++++++++++++++++
 .../LongestRepeatingCharacterReplacement.md   | 24 ++++++++
 .../editor/cn/doc/note/ContainsDuplicateIi.md | 20 +++++++
 3 files changed, 100 insertions(+)
 create mode 100644 src/main/java/leetcode/editor/cn/LongestRepeatingCharacterReplacement.java
 create mode 100644 src/main/java/leetcode/editor/cn/doc/content/LongestRepeatingCharacterReplacement.md
 create mode 100644 src/main/java/leetcode/editor/cn/doc/note/ContainsDuplicateIi.md

diff --git a/src/main/java/leetcode/editor/cn/LongestRepeatingCharacterReplacement.java b/src/main/java/leetcode/editor/cn/LongestRepeatingCharacterReplacement.java
new file mode 100644
index 0000000..c80ca9c
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/LongestRepeatingCharacterReplacement.java
@@ -0,0 +1,56 @@
+//给你一个仅由大写英文字母组成的字符串，你可以将任意位置上的字符替换成另外的字符，总共可最多替换 k 次。在执行上述操作后，找到包含重复字母的最长子串的长度。
+// 
+//
+// 注意：字符串长度 和 k 不会超过 10⁴。 
+//
+// 
+//
+// 示例 1： 
+//
+// 
+//输入：s = "ABAB", k = 2
+//输出：4
+//解释：用两个'A'替换为两个'B',反之亦然。
+// 
+//
+// 示例 2： 
+//
+// 
+//输入：s = "AABABBA", k = 1
+//输出：4
+//解释：
+//将中间的一个'A'替换为'B',字符串变为 "AABBBBA"。
+//子串 "BBBB" 有最长重复字母, 答案为 4。
+// 
+// Related Topics 哈希表 字符串 滑动窗口 👍 476 👎 0
+
+package leetcode.editor.cn;
+
+//424:替换后的最长重复字符
+class LongestRepeatingCharacterReplacement {
+    public static void main(String[] args) {
+        //测试代码
+        Solution solution = new LongestRepeatingCharacterReplacement().new Solution();
+    }
+
+    //力扣代码
+    //leetcode submit region begin(Prohibit modification and deletion)
+    class Solution {
+        public int characterReplacement(String s, int k) {
+            int[] num = new int[26];
+            int max = 0;
+            int index = 0;
+            for (int i = 0;i<s.length();i++) {
+                num[s.charAt(i) - 'A']++;
+                max = Math.max(max, num[s.charAt(i) - 'A']);
+                if (i - index + 1 - max > k) {
+                    num[s.charAt(index) - 'A']--;
+                    index++;
+                }
+            }
+            return s.length() - index;
+        }
+    }
+//leetcode submit region end(Prohibit modification and deletion)
+
+}
\ No newline at end of file
diff --git a/src/main/java/leetcode/editor/cn/doc/content/LongestRepeatingCharacterReplacement.md b/src/main/java/leetcode/editor/cn/doc/content/LongestRepeatingCharacterReplacement.md
new file mode 100644
index 0000000..e9b5413
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/doc/content/LongestRepeatingCharacterReplacement.md
@@ -0,0 +1,24 @@
+<p>给你一个仅由大写英文字母组成的字符串，你可以将任意位置上的字符替换成另外的字符，总共可最多替换 <em>k </em>次。在执行上述操作后，找到包含重复字母的最长子串的长度。</p>
+
+<p><strong>注意：</strong>字符串长度 和 <em>k </em>不会超过 10<sup>4</sup>。</p>
+
+<p> </p>
+
+<p><strong>示例 1：</strong></p>
+
+<pre>
+<strong>输入：</strong>s = "ABAB", k = 2
+<strong>输出：</strong>4
+<strong>解释：</strong>用两个'A'替换为两个'B',反之亦然。
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>s = "AABABBA", k = 1
+<strong>输出：</strong>4
+<strong>解释：</strong>
+将中间的一个'A'替换为'B',字符串变为 "AABBBBA"。
+子串 "BBBB" 有最长重复字母, 答案为 4。
+</pre>
+<div><div>Related Topics</div><div><li>哈希表</li><li>字符串</li><li>滑动窗口</li></div></div><br><div><li>👍 476</li><li>👎 0</li></div>
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diff --git a/src/main/java/leetcode/editor/cn/doc/note/ContainsDuplicateIi.md b/src/main/java/leetcode/editor/cn/doc/note/ContainsDuplicateIi.md
new file mode 100644
index 0000000..5b62de8
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/doc/note/ContainsDuplicateIi.md
@@ -0,0 +1,20 @@
+<p><strong>思路：</strong></p>
+    <p>将每个元素与其后的k个元素依次比较，看看他们是不是相等</p>
+    <p>由于题目没有给出k的范围，要注意先判断k的大小，如果k小于等于0，结果必然返回false</p>
+<p><strong>代码：</strong></p>
+<pre>
+public boolean containsNearbyDuplicate(int[] nums, int k) {
+    if (k <= 0) {
+        return false;
+    }
+    for (int i = 0; i < nums.length; i++) {
+        for (int j = i + 1; j <= Math.min(i + k, nums.length - 1); j++) {
+            if (nums[i] == nums[j]) {
+                return true;
+            }
+        }
+    }
+    return false;
+}
+</pre>
+    
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