Merge pull request apachecn#262 from ruanwenjun/leetcode-java

Add Java solution 6,7

Author: 0xkookoo

docs/Leetcode_Solutions/Java/0006._ ZigZag_Conversion.md

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+#  5. ZigZag Conversion
+
+**<font color=red>难度: Medium</font>**
+
+## 刷题内容
+
+> 原题连接
+
+* https://leetcode-cn.com/problems/zigzag-conversion/description
+
+> 内容描述
+
+```
+The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)
+
+P   A   H   N
+A P L S I I G
+Y   I   R
+
+And then read line by line: "PAHNAPLSIIGYIR"
+
+Write the code that will take a string and make this conversion given a number of rows:
+
+string convert(string s, int numRows);
+
+Example 1:
+
+Input: s = "PAYPALISHIRING", numRows = 3
+Output: "PAHNAPLSIIGYIR"
+
+
+
+Example 2:
+
+Input: s = "PAYPALISHIRING", numRows = 4
+Output: "PINALSIGYAHRPI"
+Explanation:
+
+P     I    N
+A   L S  I G
+Y A   H R
+P     I
+
+
+```
+
+## 解题方案
+
+> 思路 1
+******- 时间复杂度: O(len(s))******- 空间复杂度: O(len(s))******
+
+需要将字符串s转换为按N排列，总共有numRows行，直接将字符串转换为N字形，然后输出, beats 74.18%
+
+```java
+class Solution {
+    // 将字符串进行z子排列,行数为numRows
+    public String convert(String s, int numRows) {
+       // 思路：先转换为z字
+        List[] arr = new List[numRows]; // 保存每一行元素
+        for(int i = 0; i < numRows; i ++){
+            arr[i] = new ArrayList();
+        }
+        char[] chars = s.toCharArray();
+        for(int i = 0; i < chars.length;){
+            // 每次打印两列
+            for(int j = 0; j < numRows && i < chars.length; j++,i++){
+                List list = arr[j];
+                list.add(chars[i]);
+            }
+            for(int j = numRows - 1; j >= 0 && i < chars.length; j --){
+                if(j == numRows - 1 || j == 0){
+                    arr[j].add(' ');
+                }else{
+                    arr[j].add(chars[i]);
+                    i++;
+                }
+            }
+        }
+        // 输出最终字符串
+        char[] result = new char[chars.length];
+        int index = 0;
+        for(int i = 0; i < numRows; i ++){
+            List list = arr[i];
+            for(int j = 0; j < list.size(); j ++){
+                if(' ' != (char)list.get(j)){
+                    result[index++] = (char) list.get(j);
+                }
+            }
+        }
+        return new String(result);
+    }
+}
+```

docs/Leetcode_Solutions/Java/0007._Reverse_Integer.md

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+#  5. Reverse Integer
+
+**<font color=red>难度: Easy</font>**
+
+## 刷题内容
+
+> 原题连接
+
+* https://leetcode-cn.com/problems/reverse-integer/description
+
+> 内容描述
+
+```
+Given a 32-bit signed integer, reverse digits of an integer.
+
+Example 1:
+
+Input: 123
+Output: 321
+
+Example 2:
+
+Input: -123
+Output: -321
+
+Example 3:
+
+Input: 120
+Output: 21
+
+
+```
+
+## 解题方案
+
+> 思路 1
+******- 时间复杂度: O(n)******- 空间复杂度: O(1)******
+
+将整数翻转，翻转后是否溢出了, beats 95.35%
+
+```java
+class Solution {
+    public int reverse(int x) {
+        // 使用一个long型变量来保存
+        long index = 0;
+        while (x != 0){
+            index = index * 10 + x %10;
+            x = x / 10;
+        }
+        int result = (int) index;
+        if(result != index){
+            return 0;
+        }
+        return (int)index;
+    }
+}
+```