68 lines
1.8 KiB
Java
68 lines
1.8 KiB
Java
# 21. 调整数组顺序使奇数位于偶数前面
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[NowCoder](https://www.nowcoder.com/practice/beb5aa231adc45b2a5dcc5b62c93f593?tpId=13&tqId=11166&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
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## 题目描述
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需要保证奇数和奇数，偶数和偶数之间的相对位置不变，这和书本不太一样。
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<div align="center"> <img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/d03a2efa-ef19-4c96-97e8-ff61df8061d3.png" width="200px"> </div><br>
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## 解题思路
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方法一：创建一个新数组，时间复杂度 O(N)，空间复杂度 O(N)。
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```java
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public void reOrderArray(int[] nums) {
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// 奇数个数
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int oddCnt = 0;
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for (int x : nums)
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if (!isEven(x))
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oddCnt++;
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int[] copy = nums.clone();
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int i = 0, j = oddCnt;
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for (int num : copy) {
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if (num % 2 == 1)
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nums[i++] = num;
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else
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nums[j++] = num;
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}
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}
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private boolean isEven(int x) {
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return x % 2 == 0;
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}
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```
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方法二：使用冒泡思想，每次都当前偶数上浮到当前最右边。时间复杂度 O(N<sup>2</sup>)，空间复杂度 O(1)，时间换空间。
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```java
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public void reOrderArray(int[] nums) {
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int N = nums.length;
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for (int i = N - 1; i > 0; i--) {
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for (int j = 0; j < i; j++) {
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if (isEven(nums[j]) && !isEven(nums[j + 1])) {
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swap(nums, j, j + 1);
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}
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}
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}
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}
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private boolean isEven(int x) {
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return x % 2 == 0;
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}
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private void swap(int[] nums, int i, int j) {
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int t = nums[i];
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nums[i] = nums[j];
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nums[j] = t;
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}
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```
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<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-2.png"></img></div>
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