75. Sort Colors

栏目: Java · 发布时间: 6年前

内容简介:Given an array with n objects colored red, white or blue, sort them in-place 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

Given an array with n objects colored red, white or blue, sort them in-place 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.

Example:

Input: [2,0,2,1,1,0]
Output: [0,0,1,1,2,2]

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 a one-pass algorithm using only constant space?

Runtime: 0 ms, faster than 100.00% of Java online submissions for Sort Colors.

Memory Usage: 37.3 MB, less than 0.93% of Java online submissions for Sort Colors.

难度:medium

题目:给定一数组表示红,白,蓝三种颜色,原地 排序 使得相同颜色的物体相邻。

思路:

  1. counting sort
  2. 使用左右置换指针

counting sort

class Solution {
    public void sortColors(int[] nums) {
        int[] colors = {0, 0, 0};
        for (int i = 0; i < nums.length; i++) {
            colors[nums[i]]++;
        }
        for (int i = 0; i < nums.length;) {
            for (int j = 0; j < colors.length; j++) {
                for (int k = 0; k < colors[j]; k++) {
                    nums[i++] = j;
                }
            }
        }
    }
}

左右指针置换

class Solution {
    public void sortColors(int[] nums) {
        int left = 0;
        int right = nums.length - 1;
        for (int i = left; i <= right;) {
            if (2 == nums[i]) {
                swap(nums, i, right--);
            } else {
                i++;
            }
        }
        
        for (int i = right; i >= left;) {
            if (0 == nums[i]) {
                swap(nums, i, left++);
            } else {
                i--;
            }
        }
    }
    
    private void swap(int[] nums, int i, int j) {
        int tmp = nums[i];
        nums[i] = nums[j];
        nums[j] = tmp;
    }
}

以上就是本文的全部内容,希望对大家的学习有所帮助,也希望大家多多支持 码农网

查看所有标签

猜你喜欢:

本站部分资源来源于网络,本站转载出于传递更多信息之目的,版权归原作者或者来源机构所有,如转载稿涉及版权问题,请联系我们

智慧社会

智慧社会

阿莱克斯·彭特兰 (Alex Pentland) / 汪小帆、汪容 / 浙江人民出版社 / 2015-4 / CNY 56.90

●如果要在大数据领域推举出一个代表性的科学家,阿莱克斯·彭特兰是一个无法令人忽略的名字。经过数年极具开创性的研究,社会物理学这个全新科学领域的根基已足够深厚。社会物理学是关于想法流的科学,正是在想法流的帮助下,我们才得以提高集体智能,促进智慧社会的形成。 ● 通过研究数以百万计的人在智能手机、GPS设备、互联网等地方留下的“数字面包屑”,大数据的应用已成为一股无法被忽视的力量。在大数据的应用......一起来看看 《智慧社会》 这本书的介绍吧!

HTML 编码/解码
HTML 编码/解码

HTML 编码/解码

MD5 加密
MD5 加密

MD5 加密工具

XML、JSON 在线转换
XML、JSON 在线转换

在线XML、JSON转换工具