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✨ search in rotated sorted array

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	`1`	`+/**`
	`2`	`+ * @param {number[]} nums`
	`3`	`+ * @return {void} Do not return anything, modify nums in-place instead.`
	`4`	`+ */`
	`5`	`+function nextPermutation(nums) {`
	`6`	`+ let i = nums.length - 2; // 向左遍历,i从倒数第二开始是为了nums[i+1]要存在`
	`7`	`+ while (i >= 0 && nums[i] >= nums[i + 1]) {`
	`8`	`+ // 寻找第一个小于右邻居的数`
	`9`	`+ i--;`
	`10`	`+ }`
	`11`	`+`
	`12`	`+ if (i >= 0) {`
	`13`	`+ // 这个数在数组中存在,从它身后挑一个数,和它换`
	`14`	`+ let j = nums.length - 1; // 从最后一项,向左遍历`
	`15`	`+ while (j >= 0 && nums[j] <= nums[i]) {`
	`16`	`+ // 寻找第一个大于 nums[i] 的数`
	`17`	`+ j--;`
	`18`	`+ }`
	`19`	`+ [nums[i], nums[j]] = [nums[j], nums[i]]; // 两数交换,实现变大`
	`20`	`+ }`
	`21`	`+`
	`22`	`+ // 如果 i = -1,说明是递减排列,如 3 2 1,没有下一排列,直接翻转为最小排列:1 2 3`
	`23`	`+ let l = i + 1;`
	`24`	`+ let r = nums.length - 1;`
	`25`	`+ while (l < r) {`
	`26`	`+ // i 右边的数进行翻转,使得变大的幅度小一些`
	`27`	`+ [nums[l], nums[r]] = [nums[r], nums[l]];`
	`28`	`+ l++;`
	`29`	`+ r--;`
	`30`	`+ }`
	`31`	`+}`

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