Partition Array

Sort, Array, Two Pointers

Medium

Problem

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

Analysis

根据给定的k，也就是类似于Quick Sort中的pivot，将array从两头进行缩进，时间复杂度 O(n)

Solution

Quick Select

public class Solution {
    private void swap(int i, int j, int[] arr) {
        int tmp = arr[i];
        arr[i] = arr[j];
        arr[j] = tmp;
    }
    /**
     *@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 pl = 0;
        int pr = nums.length - 1;
        while (pl <= pr) {
            while (pl <= pr && nums[pl] < k) {
                pl++;
            }
            while (pl <= pr && nums[pr] >= k) {
                pr--;
            }
            if (pl <= pr) {
                swap(pl, pr, nums);
            }
        }
        return pl;
    }
}

Another approach: O(n)

设置一个offset值，每次将 <k 的值与offset位置上的值交换位置，然后offset++。因为不需要担心顺序，所以可以无论exchange的数是否小于k，最后所有左边的数都会小于k

public class Solution {
    /**
     * @param nums: The integer array you should partition
     * @param k: An integer
     * @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 offset = 0; 
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] < k) {
                int temp = nums[i];
                nums[i] = nums[offset]; 
                nums[offset] = temp;
                offset ++; 
            }
        }

        return offset; 
    }
}