• lintcode 中等题:partition array 数组划分


    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
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  • 原文地址:https://www.cnblogs.com/bbbblog/p/5102312.html
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