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Commit 9c0eb9d

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Merge pull request MisterBooo#90 from xiaoshuai96/master
solved @xiaoshuai96 
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## LeetCode第1054号问题:距离相等的条形码
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> 本文首发于公众号「图解面试算法」,是 [图解 LeetCode ](<https://github.com/MisterBooo/LeetCodeAnimation>) 系列文章之一。
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>
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> 同步个人博客:www.zhangxiaoshuai.fun
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**本题选自leetcode第1054号问题,medium级别,目前通过率33.3%**
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**题目描述:**
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在一个仓库里,有一排条形码,其中第 i 个条形码为 barcodes[i]。
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请你重新排列这些条形码,使其中两个相邻的条形码不能相等。
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你可以返回任何满足该要求的答案,此题保证存在答案。
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示例 1:
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输入:[1,1,1,2,2,2]
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输出:[2,1,2,1,2,1]
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示例 2:
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输入:[1,1,1,1,2,2,3,3]
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输出:[1,3,1,3,2,1,2,1]
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提示:
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1 <= barcodes.length <= 10000
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1 <= barcodes[i] <= 10000
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### 题目分析:
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1.首先我们需要将每个条形码和出现的次数作一记录,为了存取方便,这里使用数组(题目中已经给出了数组的最大和最小长度)进行操作;
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2.找出其中出现最多次数的条形码,拿到该barcode和count;
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3.先将出现次数最多的条形码存入目标数组中(偶数位或者奇数位),并对记录数组作一更新;
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4.随后将剩余的barcode填充进目标数组中。
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### GIF动画展示:
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![](../Animation/1054-rearrangeBarcodes.gif)
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### 代码:
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```java
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public static int[] rearrangeBarcodes(int[] barcodes){
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int[] address = new int[10001];
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for (int barcode : barcodes)
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address[barcode]++;
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// 找到出现次数最多的barcode
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int maxCode = 0, maxCount = 0;
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for (int i = 0; i < address.length; i++) {
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if (maxCount < address[i]) {
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maxCode = i;
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maxCount = address[i];
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}
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}
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int index = 0;
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// 先填充最大的那一位barcode
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for (; address[maxCode] > 0; index += 2) {
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barcodes[index] = maxCode;
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address[maxCode]--;
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}
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// 继续填充剩余的条形码
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for (int i = 1; i < address.length; i++) {
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while (address[i] > 0) {
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//偶数位填充完毕
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if (index >= barcodes.length) index = 1;
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barcodes[index] = i;
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address[i]--;
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index += 2;
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}
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}
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return barcodes;
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}
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```

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