您的位置:首页 > 其它

lintcode-medium-Partition Array

2016-04-03 19:03 489 查看
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

Example

If nums =
[3,2,2,1]
and
k=2
, a valid answer is
1
.

Challenge

Can you partition the array in-place and in O(n)?

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 left = 0;
int right = nums.length - 1;

while(true){
while(left < right && nums[left] < k)
left++;
while(left < right && nums[right] >= k)
right--;

if(left == right)
break;

swap(nums, left, right);
}

if(nums[left] < k)
return left + 1;
else
return left;
}

public void swap(int[] nums, int i, int j){
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
return;
}

}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: