1437. Check If All 1’s Are at Least Length K Places Away

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:

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:

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:

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

Example 4:

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

Constraints:

• 1 <= nums.length <= 10^5
• 0 <= k <= nums.length
• nums[i] is 0 or 1

---

给定一个0/1数组，问数组中所有1的间隔是否超过某个阈值。

简单题，一遍扫描数组，使用双指针记录相邻两个1，并实时计算它们的距离，完整代码如下：

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

本代码提交AC，用时140MS。