From 48fa58f8f214571d2ccbeaf6ecb176fb68360d46 Mon Sep 17 00:00:00 2001
From: dingjiawen <745518019@qq.com>
Date: Fri, 9 Sep 2022 15:02:46 +0800
Subject: [PATCH] =?UTF-8?q?leecode=E6=9B=B4=E6=96=B0?=
MIME-Version: 1.0
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---
 .../markilue/leecode/string/ReverseStr.java   | 147 ++++++++++++++++++
 .../leecode/string/ReverseString.java         |  63 ++++++++
 .../markilue/leecode/string/ReverseWords.java | 104 +++++++++++++
 3 files changed, 314 insertions(+)
 create mode 100644 Leecode/src/main/java/com/markilue/leecode/string/ReverseStr.java
 create mode 100644 Leecode/src/main/java/com/markilue/leecode/string/ReverseString.java
 create mode 100644 Leecode/src/main/java/com/markilue/leecode/string/ReverseWords.java

diff --git a/Leecode/src/main/java/com/markilue/leecode/string/ReverseStr.java b/Leecode/src/main/java/com/markilue/leecode/string/ReverseStr.java
new file mode 100644
index 0000000..50387ac
--- /dev/null
+++ b/Leecode/src/main/java/com/markilue/leecode/string/ReverseStr.java
@@ -0,0 +1,147 @@
+package com.markilue.leecode.string;
+
+import org.junit.Test;
+
+/**
+ * @BelongsProject: Leecode
+ * @BelongsPackage: com.markilue.leecode.string
+ * @Author: dingjiawen
+ * @CreateTime: 2022-09-09  10:16
+ * @Description: TODO  力扣541题  反转字符串II:
+ * 给定一个字符串 s 和一个整数 k，从字符串开头算起，每计数至 2k 个字符，就反转这 2k 字符中的前 k 个字符。
+ * 1)如果剩余字符少于 k 个，则将剩余字符全部反转。
+ * 2)如果剩余字符小于 2k 但大于或等于 k 个，则反转前 k 个字符，其余字符保持原样。
+ * @Version: 1.0
+ */
+public class ReverseStr {
+
+    @Test
+    public void test() {
+
+        String s = "abcdefghij";
+        int k = 3;
+        String s1 = reverseStr(s, k);
+        System.out.println(s1);
+
+    }
+
+    @Test
+    public void test1() {
+
+        String s = "abcd";
+        int k = 2;
+        String s1 = reverseStr(s, k);
+        System.out.println(s1);
+
+    }
+
+    @Test
+    public void test2() {
+
+        String s = "hyzqyljrnigxvdtneasepfahmtyhlohwxmkqcdfehybknvdmfrfvtbsovjbdhevlfxpdaovjgunjqlimjkfnqcqnajmebeddqsgl";
+        int k = 39;
+        String s1 = reverseStr(s, k);
+        System.out.println(s1);
+
+    }
+
+    @Test
+    public void test3() {
+
+        String s = "abcdefg";
+        int k = 39;
+        String s1 = reverseStr(s, k);
+        System.out.println(s1);
+
+    }
+
+    /**
+     * 本人的思路:fori一下，i步长为2k,2k内部就自己进行交换；for完之后判断i==length还是i<length-k还是i<length-2k做最后的处理
+     * 速度击败5.11%，内存击败8.84%
+     *
+     * @param s
+     * @param k
+     * @return
+     */
+    public String reverseStr(String s, int k) {
+
+        char[] chars = s.toCharArray();
+
+        int i = 0;
+        if (chars.length > 2 * k) {
+            //留下最后2k以内的范围单独处理,因为要分情况单独处理
+            for (; i < chars.length - 2 * k; i += 2 * k) {
+                int z = i;
+                //前k个字母反转:i和i+k-1交换
+                while (z < i + k / 2) {
+                    exchange(chars, z, i + k - 1 - z + i);
+                    z++;
+                }
+            }
+        }
+
+        //出来以后判断是从什么条件出来的
+        if (i >= chars.length - k) {
+            //剩下的长度<k
+            //有多少交换多少
+            int z = i;
+            while (z < i + (chars.length - i) / 2) {
+                exchange(chars, z, chars.length - 1 - z + i);
+                z++;
+            }
+        } else if (i >= chars.length - 2 * k) {
+            //剩下的长度<2k
+            //交换前k个
+            int z = i;
+            while (z < i + k / 2) {
+                exchange(chars, z, i + k - 1 - z + i);
+                z++;
+            }
+        }
+
+
+        String result = "";
+        for (char aChar : chars) {
+            result += aChar;
+        }
+        return result;
+    }
+
+    /**
+     * 官方的思路:与本人的类似，但是不用留下剩下的做判断,都把他视作交换全部的一部分，只不过是左右范围不同
+     * 速度击败100%，内存击败35%
+     *
+     * @param s
+     * @param k
+     * @return
+     */
+    public String reverseStr1(String s, int k) {
+
+        char[] chars = s.toCharArray();
+        //均是交换，只不过是交换的范围不同，这样就可以不用分头处理了
+        for (int i = 0; i < chars.length; i += 2 * k) {
+            reverse(chars,i,Math.min(i+k,chars.length)-1);
+        }
+
+        return new String(chars);
+
+    }
+
+    public void reverse(char[] chars,int left,int right){
+        while (left<right){
+            char temp =chars[left];
+            chars[left]=chars[right];
+            chars[right]=temp;
+            left++;
+            right--;
+        }
+    }
+
+
+    //交换数组中的两个元素
+    public void exchange(char[] s, int i, int j) {
+        char temp = s[i];
+        s[i] = s[j];
+        s[j] = temp;
+    }
+}
diff --git a/Leecode/src/main/java/com/markilue/leecode/string/ReverseString.java b/Leecode/src/main/java/com/markilue/leecode/string/ReverseString.java
new file mode 100644
index 0000000..f284b79
--- /dev/null
+++ b/Leecode/src/main/java/com/markilue/leecode/string/ReverseString.java
@@ -0,0 +1,63 @@
+package com.markilue.leecode.string;
+
+import org.junit.Test;
+
+/**
+ * @BelongsProject: Leecode
+ * @BelongsPackage: com.markilue.leecode.string
+ * @Author: dingjiawen
+ * @CreateTime: 2022-09-09  09:30
+ * @Description:
+ * TODO 力扣344题  反转字符串:
+ *  编写一个函数，其作用是将输入的字符串反转过来。输入字符串以字符数组 s 的形式给出。
+ *  不要给另外的数组分配额外的空间，你必须原地修改输入数组、使用 O(1) 的额外空间解决这一问题。
+ *
+ * @Version: 1.0
+ */
+public class ReverseString {
+
+    @Test
+    public void test(){
+        char[] s = {'h', 'e', 'l', 'l', 'o'};
+        reverseString(s);
+
+        char[] c = {'H','a','b','n','a','h'};
+        reverseString(c);
+    }
+
+    @Test
+    public void test1(){
+        char[] s = {'A',' ','m','a','n',',',' ','a',' ','p','l','a','n',',',' ','a',' ','c','a','n','a','l',':',' ','P','a','n','a','m','a'};
+        reverseString(s);
+        
+    }
+
+
+
+    /**
+     * 自己的思路:由于不让给数组分配另外的空间，因此考虑使用前后交换法，找到数组的中间然后前后交换
+     * 速度击败100% 内存击败24.24%
+     * @param s
+     */
+    public void reverseString(char[] s) {
+
+        if (s.length<=1) {
+            return;
+        }
+
+        for (int i = 0; i < s.length / 2; i++) {
+            exchange(s,i, s.length-1-i);
+        }
+//        System.out.println(s.toString());
+
+        System.out.println(s);
+
+    }
+
+    //交换数组中的两个元素
+    public void exchange(char[] s,int i,int j){
+        char temp=s[i];
+        s[i]=s[j];
+        s[j]=temp;
+    }
+}
diff --git a/Leecode/src/main/java/com/markilue/leecode/string/ReverseWords.java b/Leecode/src/main/java/com/markilue/leecode/string/ReverseWords.java
new file mode 100644
index 0000000..9205569
--- /dev/null
+++ b/Leecode/src/main/java/com/markilue/leecode/string/ReverseWords.java
@@ -0,0 +1,104 @@
+package com.markilue.leecode.string;
+
+import org.junit.Test;
+
+import java.util.Arrays;
+import java.util.Collections;
+import java.util.List;
+import java.util.stream.Stream;
+
+/**
+ * @BelongsProject: Leecode
+ * @BelongsPackage: com.markilue.leecode.string
+ * @Author: dingjiawen
+ * @CreateTime: 2022-09-09  12:12
+ * @Description: TODO  力扣151题：反转字符串中的单词:
+ * 给你一个字符串 s ，请你反转字符串中 单词 的顺序。
+ * 单词 是由非空格字符组成的字符串。s 中使用至少一个空格将字符串中的 单词 分隔开。
+ * 返回 单词 顺序颠倒且 单词 之间用单个空格连接的结果字符串。
+ * 注意：输入字符串 s中可能会存在前导空格、尾随空格或者单词间的多个空格。返回的结果字符串中，单词间应当仅用单个空格分隔，且不包含任何额外的空格。
+ * @Version: 1.0
+ */
+public class ReverseWords {
+
+    @Test
+    public void test() {
+        String s = "  hello world  ";
+        String s1 = reverseWords(s);
+        System.out.println(s1);
+    }
+
+    @Test
+    public void test1() {
+        String s = " the  sky is  blue ";
+        String s1 = reverseWords(s);
+        System.out.println(s1);
+    }
+
+
+    /**
+     * 速度超过12%，内存超过5%
+     * @param s
+     * @return
+     */
+    public String reverseWords(String s) {
+
+        String[] s1 = s.split(" ");
+        String result = "";
+        for (int i = s1.length - 1; i >= 0; i--) {
+            if (!"".equals(s1[i]) ) {
+                result += s1[i] + " ";
+            }
+        }
+
+        return result.trim();
+
+    }
+
+    /**
+     * 速度超过12%，内存超过5%
+     * @param s
+     * @return
+     */
+    public String reverseWords1(String s) {
+
+        String[] s1 = s.trim().split(" ");
+        String result = "";
+        for (int i = s1.length - 1; i >= 0; i--) {
+            if (!"".equals(s1[i]) ) {
+                if(i==0){
+                    result += s1[i];
+                }else {
+                    result += s1[i] + " ";
+                }
+
+            }
+        }
+        return result;
+
+    }
+
+    /**
+     * 官方使用API的代码
+     * 速度超过12%，内存超过5%
+     * @param s
+     * @return
+     */
+    public String reverseWords2(String s) {
+
+        // 除去开头和末尾的空白字符
+        s = s.trim();
+        // 正则匹配连续的空白字符作为分隔符分割
+        List<String> wordList = Arrays.asList(s.split("\\s+"));
+        Collections.reverse(wordList);
+        return String.join(" ", wordList);
+
+
+    }
+
+
+
+
+
+
+}