• Introduction to Algorithms


    5. 线性时间排序

      Method: 不通过比较数据进行排序,要求数组值在一定范围内;根据数组值的大小范围K,生成一个K大小的新数组,依序存储原数组的值的数量,最后取出新数组的内容即可。

     1 public class XianxingShijianPaixu {
     2     public int[] solution(int[] input, int k){
     3         int[] kElem = new int[k + 1];
     4         int n = input.length - 1;
     5         int[] output = new int[n + 1];
     6         for(int i = 0; i <= n; i++){
     7             kElem[input[i]] = kElem[input[i]] + 1;
     8         }
     9         for(int j = 1; j <= k; j++){
    10             kElem[j] = kElem[j] + kElem[j - 1];
    11         }
    12         for(int h = 0; h <= n; h++){
    13 //            System.out.println(input[h]+ " h:" + h);
    14 //            System.out.println(kElem[input[h]]+ " h:" + h);
    15 
    16             output[kElem[input[h]]-1] = input[h];
    17 //            System.out.println(input[h]+ " h:" + h);
    18             kElem[input[h]]--;
    19 //            System.out.println(kElem[input[h]]);
    20 
    21         }
    22         return output;
    23     }
    24     public static void main(String[] args){
    25 //        int[] x = {1,3,4,6,57,34,23,897,567,3,2,6,10};
    26         int[] x = {2,8,7,1,3,5,6,4,0,5,3,2};
    27 //        int[] x = {1,0,2};
    28         int[] y = new XianxingShijianPaixu().solution(x, 8);                                                                                                                                                                                                                                                                
    29 //        int[] y = new KuaiPai().secondPart(x, 0, x.length - 1);
    30         System.out.println(y[5] + " " + y[6] + " ");
    31     }
    32 }
    View Code

    4. 快速排序

      Method:FirstPart使用位置标记i 和 j,保证i 位置处之前的均为比array[array.length - 1]小,array[i + 1]设置为array[array.length - 1],其后均为比array大的项;SencondPart两次递归使用FirstPart实现最终排序。(可以随机选择array[array.length - 1]值,便于改善平均时间复杂度)

     1 import java.util.Random;
     2 
     3 public class KuaiPai {
     4     
     5     int i1;
     6     public int firstPart(int[] input, int p, int r){
     7         int i = p - 1;
     8         int tmp = 0;
     9         for(int j = p; j < r; j++){
    10             if(input[j] < input[r]){
    11                 i++;
    12                 tmp = input[j];
    13                 input[j] = input[i];
    14                 input[i] = tmp;
    15             }
    16         }
    17         tmp = input[r];
    18         input[r] = input[i + 1];
    19         input[i + 1] = tmp; 
    20         System.out.println(input[p] +" " + input[r]);
    21         return i;
    22                 
    23     }
    24     
    25     public int firstPartRandom(int[] input, int p, int r){
    26         Random random = new Random();
    27         int i = random.nextInt(r - p) + p;
    28         int tmp = input[r];
    29         input[r] = input[i];
    30         input[i] = tmp;
    31         return firstPart(input, p, r);
    32     }
    33     
    34     public int[] secondPart(int[] input, int p ,int r){
    35 //        int i = 0;
    36         if (p < r){
    37             i1 = firstPartRandom(input, p , r);
    38             System.out.println("i1: " + i1);
    39             secondPart(input, p , i1);
    40             secondPart(input, i1 + 2, r);
    41         }
    42         return input;
    43     }
    44     public static void main(String[] args){
    45         int[] x = {1,3,4,6,57,34,23,897,567,3,2,6,10};
    46 //        int[] x = {2,8,7,1,3,5,6,4};
    47 //        int[] x = {1,0,2};
    48 //        int y = new KuaiPai().firstPart(x, 0, x.length - 1 );                                                                                                                                                                                                                                                                
    49         int[] y = new KuaiPai().secondPart(x, 0, x.length - 1);
    50         System.out.println(y[2] + " " + y[12] + " ");
    51     }
    52 }
    View Code

    3. 堆排序(三步排完)

      Method:FirstPart假设左右两子堆是最大堆,与父堆比较生成新的一个最大堆,需要一次递归调用;SecondPart设置一个无序数组从Array.length - 1处开始调用F,保证生成一个最大堆;ThirdPart取出最大堆的最大值后,交换最后一值,调用F(array,0),完成过程。 this.arrayLength attention!

      1 //import java.lang.*;
      2 import java.lang.reflect.Array;
      3 public class DuiPaixu {
      4     int arrayLength;
      5     public int[] firstPart(int[] input){
      6         int n = input.length - 1;
      7         int left = 0;
      8         int right = 0;
      9         int max = 0;
     10         for(int i = 0; i < n; i++){
     11             if( 2 * i + 2 <= n){
     12                 left = input[2 * i + 1];
     13                 right = input[2 * i + 2];
     14                 if(left > input[i] && right > input[i]){
     15                     if(left < right){
     16                         max = right;
     17                         input[2 * i + 2] = input[i];
     18                         input[i] = max;
     19                         
     20                     }
     21                     else{
     22                         max = left;
     23                         input[2 * i + 1] = input[i];
     24                         input[i] = max;
     25                     }
     26                 }
     27                 else if(left > input[i]){
     28                     max = left;
     29                     input[2 * i + 1] = input[i];
     30                     input[i] = max;
     31                 }
     32                 else if(right > input[i]){
     33                     max = right;
     34                     input[2 * i + 2] = input[i];
     35                     input[i] = max;
     36                 }
     37             }
     38             if(2 * i + 2 > n && 2 * i + 1 == n){
     39                 left = input[2 * i + 1];
     40                 if(left > input[i]){
     41                     max = left;
     42                     input[2 * i + 1] = input[i];
     43                     input[i] = max;
     44                 }
     45             }
     46             
     47         }
     48         return input;
     49     }
     50     //T(n) = O(n) no better than following.
     51     
     52     public int[] firstPartBetter(int[] input, int i){
     53         int j = 2 * i + 1;
     54         int k = 2 * i + 2;
     55         int largest = 0;
     56         int tmp = 0;
     57         //具体情况具体写清晰
     58 //        this.arrayLength = input.length - 1;
     59 //        System.out.println(j + " k: " + k + " "+ arrayLength);
     60         if(k <= arrayLength){
     61             if(input[i] < input[j]){
     62                 largest = j;
     63             }
     64             else{
     65                 largest = i;
     66             }
     67             if(input[k] > input[largest]){
     68                 largest = k;
     69             }
     70 //            System.out.println(largest + " i: " + i);
     71             if(largest != i){
     72                 tmp = input[i];
     73                 input[i] = input[largest];
     74                 input[largest] = tmp;
     75                 firstPartBetter(input, largest);
     76             }
     77         }
     78         else if(k > arrayLength && j == arrayLength){
     79             if(input[i] < input[j]){
     80                 tmp = input[i];
     81                 input[i] = input[j];
     82                 input[j] = tmp;
     83             }
     84         }
     85         return input;
     86         
     87     }
     88     
     89     public int[] secondPart(int[] input){
     90 //        this.arrayLength = input.length - 1;
     91         System.out.println("test" + arrayLength);
     92         for(int i = arrayLength; i >= 0; i--){
     93 //            System.out.println(i);
     94             firstPartBetter(input, i);
     95         }
     96         return input;
     97     }
     98     
     99     public int[] thirdPart(int[] input){
    100         this.arrayLength = input.length -1;
    101         secondPart(input);
    102         int tmp = 0;
    103         for(int i = input.length - 1; i >= 0 ; i--){
    104             tmp = input[i];
    105             input[i] = input[0];
    106             input[0] = tmp;
    107             this.arrayLength = this.arrayLength - 1;
    108             firstPartBetter(input, 0);
    109         }
    110         return input;
    111     }
    112     
    113     public static void main(String[] args){
    114         int[] x = {78,34,67,89,23,2,14,6,7,3,8,8,33,3,3,1,335,7};
    115         int[] y = new DuiPaixu().thirdPart(x);
    116         System.out.println(y[2] + " " + y[3]+ " " + y[17]);
    117     }
    118 }
    View Code

    2. 归并排序(分治法)

      方法:FirstPart对左右各已排序完成部分进行整合排序,SecondPart对大的待排序数组进行分治操作,具体分解流程见下图,其它递归算法也可参照。

      

     1 public class GuibingPaixu {
     2     public int[] firstPart(int[] input, int p, int q, int r){
     3         int n1 = q - p + 1;
     4         int n2 = r - q;
     5         int k = 0;
     6         int[] input1 = new int[n1];
     7         int[] input2 = new int[n2];
     8         for(int i = 0; i < n1; i++){
     9             input1[i] = input[p + i];
    10 //            System.out.println(input1[i]+"/n");
    11         }
    12         for(int j = 0; j < n2; j++){
    13             input2[j] = input[q + j + 1];
    14         }
    15         int m = 0;
    16         int n = 0;
    17         while(k < n1 + n2){
    18             if(m < n1 && n < n2){
    19                 if(input1[m] <= input2[n]){
    20                     input[p + k] = input1[m];
    21                     m++;
    22                 }
    23                 else{
    24                     input[p + k] = input2[n];
    25                     n++;
    26                 }
    27             }
    28             else if(m >= n1 && n < n2){
    29                 input[p + k] = input2[n];
    30                 n++;
    31             }
    32             else{
    33                 input[p + k] = input1[m];
    34                 m++;
    35             }
    36             k++;
    37         }
    38         return input;
    39     }
    40     
    41     public int[] secondPart(int[] input, int p, int r){
    42         int q = 0;
    43         if(p < r){
    44             q = (p + r)/2;
    45             secondPart(input, p, q);
    46             System.out.println(q);
    47             System.out.println("r:" + r);
    48             secondPart(input, q + 1, r);
    49             System.out.println( "right" + q + 1);
    50             firstPart(input, p, q, r);
    51             System.out.println( "first" + q);
    52 
    53         }
    54         return input;
    55     }
    56     
    57     
    58     public static void main(String[] args){
    59         int[] x = {5,7,8,9,23,56,1,2,3,4,5,6,7,89};
    60         int[] x1 = {5,1,89,3,5,8,2,0,34,12,33,445,667,6,34};
    61         int[] y = new GuibingPaixu().firstPart(x, 1, 5, 13);
    62         int[] y1 = new GuibingPaixu().secondPart(x1, 0, 14);
    63         System.out.println(y1[0] + " " + y1[1] + " " + y1[2] + " " + y1[14]);
    64 //        System.out.println(y[0] + " " + y[1] + " " + y[2] + " " + y[13]);
    65     }
    66 }
    View Code

    1. 直接插入排序

      应用:属于比较排序的一种,最基础的排序方法,稳定,时间复杂度为O(n^2),空间复杂度为S(1).

      方法:设置待排数字角标为i,j = i, 把j对应数字与其前的字数逐个对比交换完成后,i++即可.

     1 public class Paixu {
     2     public int[] paixu(int[] input){
     3         int n = input.length - 1;
     4         for(int i = 1; i <= n; i++){
     5             int j = i;
     6             int tmp = 0;
     7             for(; j > 0 && input[j] < input[j - 1]; j--){
     8                 tmp = input[j - 1];
     9                 input[j - 1] = input[j];
    10                 input[j] = tmp;
    11             }
    12 
    13         }
    14         return input;
    15     }
    16     //T(n) = O(n ^ 2); S(n) = O(1) 插入排序
    17     public static void main(String[] args){
    18         int[] x = {3,3,7,8,21,1};
    19         int[] y = new Paixu().paixu(x);
    20         System.out.println(y[0] + " " + y[1]);
    21     }
    22 }
    View Code
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  • 原文地址:https://www.cnblogs.com/cenmny/p/7674270.html
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