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Partition Array

2024-08-31 07:42:25   221人阅读

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)?

 

此题是利用快速排序的思想，比较简单吧算是，但是细节需要注意一下。

有个错误解法，如下：

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) {
        int left = 0;
        int right = nums.length - 1;
        while (left < right) {
            while (left < right && nums[left] < k) {
                left++;
            }
            while (left < right && nums[right] >= k){
                right--;
            }
            if (left >= right) {
                break;
            }
            swap(nums, left, right);
            left++;
            right--;
        }
        return left;
    }
    public void swap(int[] nums, int a, int b) {
        int tmp = nums[a];
        nums[a] = nums[b];
        nums[b] = tmp;
    }
}

这种解法没有考虑到全部数字都小于target的情况，while里面的条件应该是left<=right，这样，即使循环到最后一步，left和right重合，left还要++才能返回数组的长度。

正确解法如下：

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) {
        int left = 0;
        int right = nums.length - 1;
        while (left <= right) {
            while (left <= right && nums[left] < k) {
                left++;
            }
            while (left <= right && nums[right] >= k){
                right--;
            }
            if (left > right) {
                break;
            }
            swap(nums, left, right);
            left++;
            right--;
        }
        return left;
    }
    public void swap(int[] nums, int a, int b) {
        int tmp = nums[a];
        nums[a] = nums[b];
        nums[b] = tmp;
    }
}

当然我当时的第一想法不是这样做的，while条件里面只要不越界就可以继续循环：

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) {
        int left = 0;
        int right = nums.length - 1;
        while (left < right) {
            while (left < nums.length && nums[left] < k) {
                left++;
            }
            while (right >= 0 && nums[right] >= k){
                right--;
            }
            if (left >= right) {
                break;
            }
            swap(nums, left, right);
            left++;
            right--;
        }
        return left;
    }
    public void swap(int[] nums, int a, int b) {
        int tmp = nums[a];
        nums[a] = nums[b];
        nums[b] = tmp;
    }
}

也通过了。。。有一点需要注意的是，不要把left++写到while里面，通不过，也是奇怪得很。。。

while (left < nums.length && nums[left++] < k)；

这样。。。

Partition Array


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