LintCode31 partitionArray array divided

Nums a given integer array and an integer k. Dividing the array (i.e., the mobile element in the array nums), such that:

Less than k all elements to the left
of all elements k greater than or equal to the right
to return the position of the divided array, i.e. a first position in the array i, satisfies nums [i] is greater than equal to k.

1. Note
   that you should really be divided nums array, rather than just counting the number of integers less than k, if all the elements in the array nums than k small, then return nums.length.

Sample
analysis array nums = [3,2,2,1] and k = 2, 1 is returned.

Challenges
using O (n) time complexity is divided on the array.

Conceptual understanding
long as l, r are moving too, l is the stop position of the first index that nums [i]> = k, like a general return l

Where r is discussed separately untouched or l, l where l untouched or return, where r untouched return r + 1

IX practice
will eventually stay in the position left = right

 public class Solution{
	 public int partitionArray(int[] nums,int k){
		if(nums==null||nums.length==0){
			 return -1;
		 }
		 
		 int left=0,right=nums.length-1;//这个注意是个小的逗号
		 while(left<right){
			 while(left<right&&nums[left]<k){
				 left++;
			 }
		     while(left<right&&nums[right]>=k){
				 right--;
			 }
		 if(left<right){
			 int temp=nums[right];
			 nums[left]=nums[right];
			 nums[right]=temp;
			
			 left++;
			 right--;
		}
	}
	
	if(nums[left]<k){
		return left+1;//也是可以理解  这个一个没动
	}
	//否则
	return left;
	}
 }

Reference other bloggers

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 l=0, r=nums.length-1;
        while (true) {            
            while (l<r && nums[l]<k) {
                l++;
            }
            while (l<r && nums[r]>=k) {
                r--;
            }
            if (l == r) break;
            swap(l, r, nums);
        }
        //if (l==0 && nums[l]>=k) return l;
        if (r==nums.length-1 && nums[r]<k){
        return r+1;
        }
        return l;
    }
    
    public void swap(int l, int r, int[] nums) {
        int temp = nums[l];
        nums[l] = nums[r];
        nums[r] = temp;
    }
}