lintcode-Array Partitioning

Gives an array of integers nums and an integer k. Dividing an array (that is, moving an element in an array nums) makes:

• All elements less than k are moved to the left
  
• All elements greater than or equal to K are moved to the right
  

Returns the position of the array, that is, the first position in the array I, satisfies nums[i] is greater than or equal to K.

Have you ever encountered this problem in a real interview? YesSample Example

give the array nums= [3,2,2,1] and k=2, return 1

Note

You should really divide the array nums, not just the number of integers smaller than k, if all the elements in the array nums are smaller than K, the Nums.length is returned.

challenges

Requires that the time complexity of O (n) be used in situ to divide the array

label Expand

Related TopicsExpand

Analysis: The principle is the partition of the fast line

Code:

Class Solution {public:    int Partitionarray (vector<int> &nums, int. k) {        //write your code here        int P = 0;        for (int i=0;i<nums.size (); i++)        {            if (nums[i]<k)            {                swap (nums[i],nums[p]);                p++;            }        }        return p;    }};

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lintcode-Array Partitioning