题目

数组划分 

给出一个整数数组nums和一个整数k。划分数组(即移动数组nums中的元素),使得:

  • 所有小于k的元素移到左边
  • 所有大于等于k的元素移到右边

返回数组划分的位置,即数组中第一个位置i,满足nums[i]大于等于k。

样例

给出数组nums=[3,2,2,1]和 k=2,返回 1

注意

你应该真正的划分数组nums,而不仅仅只是计算比k小的整数数,如果数组nums中的所有元素都比k小,则返回nums.length。

挑战

要求在原地使用O(n)的时间复杂度来划分数组

解题

快速排序,搞了好久,中间的那个值一直找不到,然后就遍历数组找了。

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.length==0)
            return 0;
        int p = partition(nums,0,nums.length-1,k);
        if(p==nums.length - 1)
            return nums.length;
        return p;
        
    }
    public int partition(int[] nums,int left,int right,int k){
        int i = left;
        int j = right;
        while(i<j){
            while(i<j&&nums[j] >=k) j--;
            
            while(i<j&& nums[i]<k) i++;
            if(nums[i]>=k && nums[j]<k){
                int tmp = nums[i];
                nums[i] = nums[j];
                nums[j] = tmp;
                j--;
                i++;
            }
        }
        for(i=0;i<right;i++)
            if(nums[i]>=k)
                return i;
        return right;
    }
}
Java Code

九章算法中,i<j 换成i<=j 最后输出的i就是所求的答案

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.length==0)
            return 0;
        int p = partition(nums,0,nums.length-1,k);
        return p;
        
    }
    public int partition(int[] nums,int left,int right,int k){
        int i = left;
        int j = right;
        while(i<=j){
            while(i<=j&&nums[j] >=k) j--;
            while(i<=j&& nums[i]<k) i++;
            if(i<j){
                int tmp = nums[i];
                nums[i] = nums[j];
                nums[j] = tmp;
                j--;
                i++;
            }
        }
        return i;
    }
}
Java Code
class Solution:
    """
    @param nums: The integer array you should partition
    @param k: As description
    @return: The index after partition
    """
    def partitionArray(self, nums, k):
        # write your code here
        # you should partition the nums by k
        # and return the partition index as description
        l = len(nums)
        if l == 0:
            return 0
        i = 0
        j = l-1
        while i<j:
            while i<=j and nums[j]>=k:
                j-=1
            while i<=j and nums[i]<k:
                i+=1
            if i<j:
                tmp = nums[i]
                nums[i]=nums[j]
                nums[j]=tmp
                i+=1
                j-=1
        return i 
Python Code