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feat: add leetcode question 198

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	`1`	`+package com.hi.dhl.algorithms.leetcode._189.java;`
	`2`	`+`
	`3`	`+/**`
	`4`	`+ * <pre>`
	`5`	`+ * author: dhl`
	`6`	`+ * date : 2022年10月30日`
	`7`	`+ * desc :`
	`8`	`+ * </pre>`
	`9`	`+ */`
	`10`	`+class Solution {`
	`11`	`+ // 方法一`
	`12`	`+ public void rotate(int[] nums, int k) {`
	`13`	`+ int len = nums.length;`
	`14`	`+ k = k % len;`
	`15`	`+ int times = 1;`
	`16`	`+ // 数组的总交换次数是和数组长度相等的`
	`17`	`+ // 所以记录一下总的交换次数即可`
	`18`	`+ for (int i = 0; i < times; i++) {`
	`19`	`+ int pre = nums[i];`
	`20`	`+ int cur = i;`
	`21`	`+ int count = 0; // 记录每次循环的交换次数`
	`22`	`+ do {`
	`23`	`+ cur = (cur + k) % len;`
	`24`	`+ int holder = nums[cur];`
	`25`	`+ nums[cur] = pre;`
	`26`	`+ pre = holder;`
	`27`	`+ count++;`
	`28`	`+ } while (i != cur); // 是否回到原点`
	`29`	`+ times = len / count; // 剩余交换次数`
	`30`	`+ }`
	`31`	`+ }`
	`32`	`+`
	`33`	`+ // 方法二`
	`34`	`+ public void rotate2(int[] nums, int k) {`
	`35`	`+ int len = nums.length;`
	`36`	`+ k = k % len;`
	`37`	`+ swap(nums, 0, len - 1);`
	`38`	`+ swap(nums, 0, k - 1);`
	`39`	`+ swap(nums, k, len - 1);`
	`40`	`+ }`
	`41`	`+`
	`42`	`+ public void swap(int[] nums, int lo, int hi) {`
	`43`	`+ while (lo < hi) {`
	`44`	`+ int holder = nums[lo];`
	`45`	`+ nums[lo] = nums[hi];`
	`46`	`+ nums[hi] = holder;`
	`47`	`+ lo++;`
	`48`	`+ hi--;`
	`49`	`+ }`
	`50`	`+ }`
	`51`	`+}`

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