Partition Array

Source

• lintcode: 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.

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)?

Python

class Solution:
    """
    @param nums: The integer array you should partition
    @param k: As description
    @return: The index after partition
    """
    def partitionArray(self, nums, k):
        # write your code here
        # you should partition the nums by k
        # and return the partition index as description
        start=0
        end=len(nums)-1;

        while start<=end:
            if nums[start]<k:
                start+=1
            else:
                nums[start], nums[end] = nums[end], nums[start]
                end-=1
        return start

Java

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) {
        int start=0;
        int end=nums.length-1;

        while(start<=end){
            if(nums[start]<k){
                start++;
            }
            else{
                swap(nums, start, end);
                end--;
            }
        }
        return start;
    }

    public void swap(int[] nums, int left, int right){
        int tmp = nums[left];
        nums[left] = nums[right];
        nums[right] = tmp;
    }
}