From f69d545dada0c6b7cd8d472e2856e6fa04d360bb Mon Sep 17 00:00:00 2001
From: cp0511 <chengpei20083152@126.com>
Date: Tue, 8 Dec 2020 09:10:28 +0800
Subject: [PATCH] =?UTF-8?q?=E7=A8=8B=E6=B2=9B-238-=E7=A7=BB=E5=8A=A8?=
 =?UTF-8?q?=E9=9B=B6?=
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---
 Week_01/cp/LeetCode_283.java | 54 ++++++++++++++++++++++++++++++++++++
 1 file changed, 54 insertions(+)

diff --git a/Week_01/cp/LeetCode_283.java b/Week_01/cp/LeetCode_283.java
index 45460b4..40292be 100644
--- a/Week_01/cp/LeetCode_283.java
+++ b/Week_01/cp/LeetCode_283.java
@@ -1,7 +1,61 @@
 package cp;
 
 /**
+ *
  * Created by 程程有小棉被啊 on 2020年12月07日.
  */
 public class LeetCode_283 {
+
+
+    /*
+    移动零
+        给定一个数组 nums，编写一个函数将所有 0 移动到数组的末尾，同时保持非零元素的相对顺序。
+        说明:
+            1.必须在原数组上操作，不能拷贝额外的数组。
+            2.尽量减少操作次数。
+        示例：
+            输入: [0,1,0,3,12]
+            输出: [1,3,12,0,0]
+     */
+    public static void main(String[] args) {
+        int[] num = {0,1,0,3,12};
+//        moveZero(num);
+        moveZero2(num);
+    }
+
+    static void moveZero(int[] num){
+        //双指针
+        int j = 0;//记录不为0的位置
+        for (int i = 0; i < num.length; i++) {
+            if (num[i] != 0){
+                num[j] = num[i];//交换位置
+                if (j != i){
+                    num[i] = 0;//将该处置为零
+                }
+                j++;
+            }
+        }
+        moveZeroPrintln(num);
+    }
+
+    static void moveZero2(int[] num){
+        //先将不是零的数字往前挪动，最后将最后的补零
+        int j =0;
+        for (int i = 0; i < num.length; i++) {
+            if (num[i] != 0){
+                num[j] = num[i];
+                j++;
+            }
+        }
+        for (int i = j; i < num.length; i++) {
+            num[i] = 0;
+        }
+        moveZero(num);
+    }
+
+    static void moveZeroPrintln(int[] num){
+        for (int i = 0; i < num.length; i++) {
+            System.out.println(num[i]);
+        }
+    }
 }
-- 
Gitee