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 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)?

Tags Expand 

Two Pointers Sort Array

Just like partition function for quick sort.
Use two pointers left and right to compare with k. If left < k, left ++; If right > k, right --. When left and right stop forwarding, exchange values. When left > right, all elements have been partitioned.
Time: O(n)
Space: O(1)

class Solution {
public:
    int partitionArray(vector<int> &nums, int k) {
        // write your code here
        int left = 0;
        int right = nums.size() - 1;
        while(left < right){
            while(nums[left] < k && left < nums.size()) left++;
            while(nums[right] >= k && right >= 0) right--;
            if(left < right){
                int temp = nums[left];
                nums[left] = nums[right];
                nums[right] = temp;
            }
        }
        //return left;
        return right + 1;
    }
};

http://www.code123.cc/docs/leetcode-notes/integer_array/partition_array.html


Lintcode second:

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23

class Solution { public: int partitionArray(vector<int> &nums, int k) { int left = 0; int right = nums.size() - 1; while(left < right){ while(nums[left] < k && left < nums.size()){ left++; } while(nums[right] >= k && right >= 0){ right--; } if(left < right){ int temp = nums[left]; nums[left] = nums[right]; nums[right] = temp; left++; right--; } } return left; } };