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Commit d1d4370

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feat: add question
1 parent a6e5276 commit d1d4370

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3 files changed

+98
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3 files changed

+98
-5
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‎00-code(源代码)/src/com/hi/dhl/algorithms/offer/_21/java/Solution.java

Lines changed: 33 additions & 3 deletions
Original file line numberDiff line numberDiff line change
@@ -50,12 +50,42 @@ public int[] exchange2(int[] nums) {
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return nums;
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}
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/**
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* 基数在前面,偶数在后面,且相对位置不变
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*
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* @return
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*/
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public int[] exchange3(int[] nums) {
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for (int i = 0; i < nums.length; i++) {
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if (nums[i] % 2 != 0) continue;
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int temp = 0;
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int evenIndex = -1;
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for (int j = i + 1; j < nums.length; j++) {
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if (nums[j] % 2 != 0) {
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temp = nums[j];
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evenIndex = j;
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break;
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}
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}
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for (int k = evenIndex; k > i; k--) {
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nums[k] = nums[k - 1];
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}
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if (temp != 0)
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nums[i] = temp;
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}
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return nums;
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}
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public static void main(String... agrs) {
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Solution sort = new Solution();
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int[] nums = new int[]{1, 2, 5, 6};
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sort.exchange(nums);
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int[] nums = new int[]{1, 6, 3, 2, 5, 4, 7, 8, 9};
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sort.exchange3(nums);
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for (int item : nums) {
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System.out.println(item);
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System.out.print(item + ", ");
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}
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}
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‎offer/_sidebar.md

Lines changed: 2 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -6,8 +6,8 @@
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* 多线程
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9-
<!-- * [阿里-线程交替打印数字](/multi-thread/01-printThread.md)
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* [头条-线程交替打印 ABC](/multi-thread/02-printABCThread.md)-->
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* [阿里-线程交替打印数字](/multi-thread/01-printThread.md)
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* [头条-线程交替打印 ABC](/multi-thread/02-printABCThread.md)
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* [多线程-按序打印](/multi-thread/03-print-in-order.md)
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* [多线程-交替打印FooBar](/multi-thread/04-print-alternately.md)
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* [多线程-打印零与奇偶数](/multi-thread/05-print-zero-even-odd.md)

‎offer/algorithm/21-diao-zheng-shu-zu-shun-xu-shi-qi-shu-wei-yu-ou-shu-qian-mian-lcof.md

Lines changed: 63 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -219,3 +219,66 @@ public class Solution {
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<!-- tabs:end -->
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### 拓展题目
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在上面题目中增加一点难度,题目如下:
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输入一个整数数组,实现一个函数来调整该数组中数字的顺序,使得所有奇数位于数组的前半部分,所有偶数位于数组的后半部分,且相对位置不变
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**示例 1:**
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```
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输入:nums = [1, 6, 3, 2, 5, 4, 7, 8, 9]
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输出:[1, 3, 5, 7, 9, 6, 2, 4, 8]
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```
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这道题目类似于插入排序,我们先看一下插入排序算法
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```
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void insertSort(int[] nums) {
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for (int i = 1; i < nums.length; i++) {
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int temp = nums[i];
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int j = i - 1;
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while (j >= 0 && nums[j] > temp) {
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nums[j + 1] = nums[j];
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j--;
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}
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nums[j + 1] = temp;
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}
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}
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```
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这道题目重点在于「 奇数位于偶数前面,**且相对位置不变** 」,核心思路如下:
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* 遍历数组,记录第一次出现偶数的位置,记为 i
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* 接着从上一次偶数位置,继续遍历寻找奇数出现的位置,记录当前下标 j 和元素 temp
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* 移动数组 [i+1,j] 范围内的元素,统一向右移动 1 位
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* 将元素 temp 放到下标为 i 的位置上
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```
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public int[] exchange(int[] nums) {
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for (int i = 0; i < nums.length; i++) {
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if (nums[i] % 2 != 0) continue;
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int temp = 0;
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int evenIndex = -1;
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for (int j = i + 1; j < nums.length; j++) {
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if (nums[j] % 2 != 0) {
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temp = nums[j];
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evenIndex = j;
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break;
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}
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}
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for (int k = evenIndex; k > i; k--) {
275+
nums[k] = nums[k - 1];
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}
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if (temp != 0)
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nums[i] = temp;
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}
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return nums;
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}
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```
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