From 5a991ac39adb0b365655ed74dd8bdbd0b29662cd Mon Sep 17 00:00:00 2001
From: liangjinglin <liangjinglin@hc-smart.com>
Date: Sun, 16 Feb 2025 17:23:44 +0800
Subject: [PATCH] =?UTF-8?q?2025-02-16=20=E7=A7=BB=E5=8A=A8=E9=9B=B6?=
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---
 src/dataANDcalc/java/leetcode/MoveZeroes.java | 60 +++++++++++++++++++
 1 file changed, 60 insertions(+)
 create mode 100644 src/dataANDcalc/java/leetcode/MoveZeroes.java

diff --git a/src/dataANDcalc/java/leetcode/MoveZeroes.java b/src/dataANDcalc/java/leetcode/MoveZeroes.java
new file mode 100644
index 0000000..2e178b0
--- /dev/null
+++ b/src/dataANDcalc/java/leetcode/MoveZeroes.java
@@ -0,0 +1,60 @@
+package leetcode;
+
+//
+//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。
+//
+//        请注意 ，必须在不复制数组的情况下原地对数组进行操作。
+//
+//
+//
+//        示例 1:
+//
+//        输入: nums = [0,1,0,3,12]
+//        输出: [1,3,12,0,0]
+//        示例 2:
+//
+//        输入: nums = [0]
+//        输出: [0]
+//
+//
+//        提示:
+//
+//        1 <= nums.length <= 104
+//        -231 <= nums[i] <= 231 - 1
+//
+//
+//        进阶：你能尽量减少完成的操作次数吗？
+
+public class MoveZeroes {
+
+
+    private void move(int[] nums){
+        int left = 0;
+        int right = 0;
+        while(true) {
+            if(right == nums.length){
+                break;
+            }
+            if(nums[left] == 0){
+                if(nums[right] != 0){
+                    int temp = nums[left];
+                    nums[left] = nums[right];
+                    nums[right] = temp;
+                    left ++;
+                }
+            } else {
+                left ++;
+            }
+            right ++;
+        }
+    }
+
+    public static void main(String[] args) {
+        int[] nums = {1,0,1};
+        MoveZeroes moveZeroes = new MoveZeroes();
+        moveZeroes.move(nums);
+        for(int num : nums){
+            System.out.print(num + " ");
+        }
+    }
+}