1752. Check if Array Is Sorted and Rotated

Easy

題目描述

Description

Given an array nums, return true if the array was originally sorted in non-decreasing order, then rotated some number of positions (including zero). Otherwise, return false.

There may be duplicates in the original array.

Note: An array A rotated by x positions results in an array B of the same length such that A[i] == B[(i+x) % A.length], where % is the modulo operation.

Example 1:

Input: nums = [3,4,5,1,2] Output: true Explanation: [1,2,3,4,5] is the original sorted array. You can rotate the array by x = 3 positions to begin on the the element of value 3: [3,4,5,1,2].

Example 2:

Input: nums = [2,1,3,4] Output: false Explanation: There is no sorted array once rotated that can make nums.

Example 3:

Input: nums = [1,2,3] Output: true Explanation: [1,2,3] is the original sorted array. You can rotate the array by x = 0 positions (i.e. no rotation) to make nums.

Constraints:

1 <= nums.length <= 100
1 <= nums[i] <= 100

Follow up: Suppose there are lots of incoming s, say s1, s2, ..., sk where k >= 109, and you want to check one by one to see if t has its subsequence. In this scenario, how would you change your code?

解答

Solution

/**
 * @param {number[]} nums
 * @return {boolean}
 */
var check = function (nums) {
  let count = 0;

  for (i = 0; i < nums.length; i++) {
    if (nums[i] > nums[(i + 1) % nums.length]) {
      count++;
    }
  }

  return count <= 1;
};

解題思路

想法就是暴力解的遍歷陣列比較陣列是否為非遞減（[3,3,3,3]這種全部都同一個數的也算非遞減所以會是 true），但由於陣列可能是旋轉過的所以有1次的寬限，如果不只1次的話就要算 false 了

心得

U點小難，一開始沒看懂題目 Rotated 的意思，想說為何[2,1]也算true，後來再讀一次之後把邏輯打掉重寫一次才成功