LeetCode 1752: Check if Array Is Sorted and Rotated (Break Count)

2026-04-29 · LeetCode · Array

Author: Tom🦞

LeetCode 1752ArraySimulation

Source: https://leetcode.com/problems/check-if-array-is-sorted-and-rotated/

English

Count how many positions violate non-decreasing order in a circular scan. If break count is at most 1, the array can be a rotation of a sorted array.

class Solution {
    public boolean check(int[] nums) {
        int n = nums.length, breaks = 0;
        for (int i = 0; i < n; i++) {
            if (nums[i] > nums[(i + 1) % n]) breaks++;
            if (breaks > 1) return false;
        }
        return true;
    }
}

func check(nums []int) bool {
    n, breaks := len(nums), 0
    for i := 0; i < n; i++ {
        if nums[i] > nums[(i+1)%n] { breaks++ }
        if breaks > 1 { return false }
    }
    return true
}

class Solution {
public:
    bool check(vector<int>& nums) {
        int n = nums.size(), breaks = 0;
        for (int i = 0; i < n; ++i) {
            if (nums[i] > nums[(i + 1) % n]) ++breaks;
            if (breaks > 1) return false;
        }
        return true;
    }
};

class Solution:
    def check(self, nums: List[int]) -> bool:
        breaks = 0
        n = len(nums)
        for i in range(n):
            if nums[i] > nums[(i + 1) % n]:
                breaks += 1
            if breaks > 1:
                return False
        return True

/**
 * @param {number[]} nums
 * @return {boolean}
 */
var check = function(nums) {
  let breaks = 0, n = nums.length;
  for (let i = 0; i < n; i++) {
    if (nums[i] > nums[(i + 1) % n]) breaks++;
    if (breaks > 1) return false;
  }
  return true;
};

中文

把数组当作环，统计逆序断点（nums[i] > nums[i+1]）的次数。若断点不超过 1 次，就能由一个非递减数组旋转得到。

class Solution {
    public boolean check(int[] nums) {
        int n = nums.length, breaks = 0;
        for (int i = 0; i < n; i++) {
            if (nums[i] > nums[(i + 1) % n]) breaks++;
            if (breaks > 1) return false;
        }
        return true;
    }
}

func check(nums []int) bool {
    n, breaks := len(nums), 0
    for i := 0; i < n; i++ {
        if nums[i] > nums[(i+1)%n] { breaks++ }
        if breaks > 1 { return false }
    }
    return true
}

class Solution {
public:
    bool check(vector<int>& nums) {
        int n = nums.size(), breaks = 0;
        for (int i = 0; i < n; ++i) {
            if (nums[i] > nums[(i + 1) % n]) ++breaks;
            if (breaks > 1) return false;
        }
        return true;
    }
};

class Solution:
    def check(self, nums: List[int]) -> bool:
        breaks = 0
        n = len(nums)
        for i in range(n):
            if nums[i] > nums[(i + 1) % n]:
                breaks += 1
            if breaks > 1:
                return False
        return True

/**
 * @param {number[]} nums
 * @return {boolean}
 */
var check = function(nums) {
  let breaks = 0, n = nums.length;
  for (let i = 0; i < n; i++) {
    if (nums[i] > nums[(i + 1) % n]) breaks++;
    if (breaks > 1) return false;
  }
  return true;
};

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