LeetCode解题报告（278）-- 1437. Check If All 1's Are at Least Length K Places Away

Problem

Given an array nums of 0s and 1s and an integer k, return True if all 1’s are at least k places away from each other, otherwise return False.

Example 1:

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2
3

Input: nums = [1,0,0,0,1,0,0,1], k = 2
Output: true
Explanation: Each of the 1s are at least 2 places away from each other.

Example 2:

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3

Input: nums = [1,0,0,1,0,1], k = 2
Output: false
Explanation: The second 1 and third 1 are only one apart from each other.

Example 3:

1 2	Input: nums = [1,1,1,1,1], k = 0 Output: true

Example 4:

1 2	Input: nums = [0,1,0,1], k = 1 Output: true

Constraints:

1 <= nums.length <= 105
0 <= k <= nums.length
nums[i] is 0 or 1

Analysis

题目给出一个只包含0和1的数组，还有一个间隔k，问这个数组中相邻的1之间的距离是否都至少为k。思路很简单，直接一遍遍历，用一个下标记录上一次出现1的位置，然后找到1，下标做差就能计算出之间的距离。

Solution

无

Code

class Solution {
public:
    bool kLengthApart(vector<int>& nums, int k) {
        int idx = -1;
        int size = nums.size();
        for (int i = 0; i < size; i++) {
            if (idx == -1 && nums[i] == 1) {
                idx = i;
            } else if (nums[i] == 1) {
                if (i - idx <= k) {
                    return false;
                } else {
                    idx = i;
                }
            }
        }
        return true;
    }
};

Summary

这是一道非常简单的数组类题目，仅在这里做一下总结，没有过多的算法知识分享。这道题这道题目的分享到这里，谢谢您的支持！