Given an array of integers nums and a positive integer k, find whether it's possible to divide this array into sets of k consecutive numbers
Return True if its possible otherwise return False.
Example 1:
Input: nums = [1,2,3,3,4,4,5,6], k = 4 Output: true Explanation: Array can be divided into [1,2,3,4] and [3,4,5,6].
Example 2:
Input: nums = [3,2,1,2,3,4,3,4,5,9,10,11], k = 3 Output: true Explanation: Array can be divided into [1,2,3] , [2,3,4] , [3,4,5] and [9,10,11].
Example 3:
Input: nums = [3,3,2,2,1,1], k = 3 Output: true
Example 4:
Input: nums = [1,2,3,4], k = 3 Output: false Explanation: Each array should be divided in subarrays of size 3.
Constraints:
1 <= nums.length <= 10^51 <= nums[i] <= 10^91 <= k <= nums.length
class Solution {
public boolean isPossibleDivide(int[] nums, int k) {
boolean res = false;
int l = nums.length;
if(l % k != 0) return res;
Map<Integer, Integer> map = new HashMap();
PriorityQueue<Integer> q = new PriorityQueue();
for(int i: nums){
map.put(i, map.getOrDefault(i, 0)+1);
}
for(int i: map.keySet()) q.offer(i);
while(!q.isEmpty()){
int cur = q.poll();
if(map.get(cur) == 0) continue;
int times = map.get(cur);
for(int i = 0; i < k; i++){
if(!map.containsKey(cur + i) || map.get(cur + i) < times) return false;
map.put(cur + i, map.get(cur + i) - times);
}
l -= k * times;
}
return l == 0;
}
}
首先判断有没有k倍的元素,然后用map记录每个数出现的数量,然后用pq记下出现的数字,并用全局变量l记录数组长度。
先拿出一个key,如果对应的value==0就跳过,否则就用times出现的数量。
从0到k-1循环,如果没有cur+i或者cur+i剩余数量少于times说明必不可能成功,return false,更新map中每个key的value
更新剩余长度l
判断l==0?
来源:https://www.cnblogs.com/wentiliangkaihua/p/12208111.html