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 rightReturn the partitioning Index, i.e the first index "i" nums[i] >= k.NoteYou 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"ExampleIf nums=[3,2,2,1] and k=2, a valid answer is 1.ChallengeCan you partition the array in-place and in O(n)?

Quick Sort 一樣的做法，只是有兩種情況特殊處理：我第一次做的時候沒有考慮到

1. all elements in nums are greater than or equal to k, l pointer never shift， should return r

2.all elements in nums are smaller than k, r pointer never shift, shoud return r+1

 1 public class Solution { 2     /**  3      *@param nums: The integer array you should partition 4      *@param k: As description 5      *return: The index after partition 6      */ 7     public int partitionArray(ArrayList<Integer> nums, int k) { 8         //write your code here 9         if (nums==null || nums.size()==0) return 0;10         int l=0, r=nums.size()-1;11         while (true) {12             while (l<r && nums.get(r)>=k) {13                 r--;14             }15             while (l<r && nums.get(l)<k) {16                 l++;17             }18             if (l == r) break;19             swap(l, r, nums);20         }21         if (l==0 && nums.get(l)>=k) return r;22         if (r==nums.size()-1 && nums.get(l)<k) return r+1;23         return r+1;24     }25     26     public void swap(int l, int r, ArrayList<Integer> nums) {27         int temp = nums.get(l);28         nums.set(l, nums.get(r).intValue());29         nums.set(r, temp);30     }31 }