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

class Solution {
    public boolean kLengthApart(int[] nums, int k) {
        boolean res = false;
        List<Integer> list = new ArrayList();
        for(int i = 0; i < nums.length; i++){
            if(nums[i] == 1) list.add(i);
        }
        System.out.println(list.toString());
        //if(list.size() == 0) return k==0;
        int st = Integer.MAX_VALUE;
        for(int i = 1; i < list.size(); i++){
            st = Math.min(st, list.get(i) - list.get(i - 1));
            if(st - 1 < k) return false;
        }
        return true;
    }
}

求两1点间最小距离，注意如果没有1直接返回true

class Solution {
    public boolean kLengthApart(int[] nums, int k) {
        for(int start = -1, i = 0; i < nums.length; i++) 
            if(nums[i] == 1) {
                if(start != -1 && i - start - 1 < k) return false;
                start = i;
            }
        return true;
    }
}

或者这样O（N）