Given an array
nums
of integers and an int k
, partition the array (i.e move the elements in "nums") such that:- All elements < k are moved to the left
- All elements >= k are moved to the right
Return the partitioning index, i.e the first index inums[i] >= k.
Have you met this question in a real interview?
Yes
Example
If nums =
[3,2,2,1]
and k=2
, a valid answer is 1
.
Note
You should do really partition in array nums instead of just counting the numbers of integers smaller than k.
If all elements in nums are smaller than k, then return nums.length
Challenge
Can you partition the array in-place and in O(n)?
public class Solution {
/**
*@param nums: The integer array you should partition
*@param k: As description
*return: The index after partition
*/
public int partitionArray(int[] nums, int k) {
//write your code here
int less = 0, great = nums.length-1;
while(less < great){
while(less <= great && nums[less] < k){
less++;
}
while(great >= less && nums[great] >= k){
great--;
}
if(less <= great){
int temp = nums[less];
nums[less] = nums[great];
nums[great] = temp;
less++;
great--;
}
}
return less;
}
}
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