zoukankan      html  css  js  c++  java
  • leetCode 75.Sort Colors (颜色排序) 解题思路和方法

    Given an array with n objects colored red, white or blue, sort them so that objects of the same color are adjacent, with the colors in the order red, white and blue.

    Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively.

    Note:
    You are not suppose to use the library's sort function for this problem.

    click to show follow up.

    Follow up:
    A rather straight forward solution is a two-pass algorithm using counting sort.
    First, iterate the array counting number of 0's, 1's, and 2's, then overwrite array with total number of 0's, then 1's and followed by 2's.

    Could you come up with an one-pass algorithm using only constant space?

    思路:假设空间上不做要求,这题还是比較简单的。能够再声明一个m,n矩阵,然后搜索原矩阵。是0。置于首位,是2置于末尾,中间置为1.

    代码例如以下:

    public class Solution {
        public void sortColors(int[] nums) {
            
            int[] a = new int[nums.length];
            a = Arrays.copyOf(nums, nums.length);
            
            int i = 0;
            int j = nums.length - 1;
            for(int k = 0; k < nums.length; k++){
                if(a[k] == 0){
                    nums[i++] = a[k];
                }else if(a[k] == 2){
                    nums[j--] = a[k];
                }
            }
            while(i <= j){
                nums[i++] = 1;
            }
        }
    }

    可是假设常数空间的话,就有点难度,网上通用的解法例如以下(能够对n个数排序):

    public class Solution {
        public void sortColors(int[] nums) {
            int i = -1,j = -1,k = -1;
            for(int m = 0; m < nums.length; m++){
                if(nums[m] == 0){
                    nums[++k] = 2;
                    nums[++j] = 1;
                    nums[++i] = 0;
                }else if(nums[m] == 1){
                    nums[++k] = 2;
                    nums[++j] = 1;
                }else{
                    nums[++k] = 2;
                }
            }
        }
    }


  • 相关阅读:
    自定义jquery插件
    jquery中的编程范式,即jquery的牛逼之处
    $.ajax 完整参数
    URL参数获取/转码
    hello world
    此博客已不更新,作者的个人域名LIZHONGC.COM已经启用。
    岁月记录
    下雪往事
    《x86汇编语言:从实模式到保护模式》检测点和习题答案
    《穿越计算机的迷雾》第二版再版说明
  • 原文地址:https://www.cnblogs.com/wzjhoutai/p/6762576.html
Copyright © 2011-2022 走看看