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 一样的做法，只是有两种情况特殊处理：我第一次做的时候没有考虑到

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

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

第一次做法（稍次）

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
     
       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
      
       =k) {13                 r--;14             }15             while (l
       
        =k) return r;22         if (r==nums.size()-1 && nums.get(l)
        
          nums) {27         int temp = nums.get(l);28         nums.set(l, nums.get(r).intValue());29         nums.set(r, temp);30     }31 }
        
       
      
     

 第二次做法（推荐）： 只要l,r 都动过，l停的位置就是first index that nums[i] >= k, 一般情况return l就好了

单独讨论l或者r没有动过的情况，l没有动过的情况还是return l, r没有动过的情况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(int[] nums, int k) { 8         //write your code here 9         if (nums==null || nums.length==0) return 0;10         int l=0, r=nums.length-1;11         while (true) {            12             while (l
     
      =k) {16                 r--;17             }18             if (l == r) break;19             swap(l, r, nums);20         }21         //if (l==0 && nums[l]>=k) return l;22         if (r==nums.length-1 && nums[r]