lintcode-medium-Partition Array

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 i nums[i] >= k.

 

 Notice

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

Example

If nums = [3,2,2,1] and k=2, a valid answer is 1.

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
        if(nums == null || nums.length == 0)
            return 0;
        
        int left = 0;
        int right = nums.length - 1;
        
        while(true){
            while(left < right && nums[left] < k)
                left++;
            while(left < right && nums[right] >= k)
                right--;
            
            if(left == right)
                break;
            
            swap(nums, left, right);
        }
        
        if(nums[left] < k)
            return left + 1;
        else
            return left;
    }
    
    public void swap(int[] nums, int i, int j){
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
        return;
    }
    
}