数组元素的移动(删除) C#实现

下面有四个问题:

  1. 把数组元素前后部分交换 MoveFirstPartOfArrayToTheEnd(int[] array, int index) 比如 {1,2,3,4,5,6,7} 3  => {4,5,6,7,1,2,3}
  2. 把数组元素前后部分交换 MoveFirstPartOfArrayToTheEnd(int[]  array, int value）比如 {1,2,8,4,5,6,7} 8  => {4,5,6,7,1,2,8}
  3. 把数组一段移动到后面MoveSomeElementsToTheEnd(int[]  array, int startIndex, int length)比如{1,2,3,4,5,6,7,8}  3  3  => {1,2,3,7,8,4,5,6}
  4. 把数组中重复的元素变成0放到最后面RemoveDulplicatedElements(int[] array) 比如 {1,3,3,2,4,4,4,5} => {1,3,2,4,5,0,0,0}

你首先想到的办法是什么？

  1. 申请一个临时数组把FistPart和EndPart交换
  2. 同上，只要找到对应值的下标即可。
  3. 同上申请临时数组，把要移动的段放到临时数组里
  4. 申请一个临时的List<int>把唯一的元素加到List里面，再重新赋给Array。
  1     class Program
  2     {
  3         static void Main(string[] args)
  4         {
  5             //1
  6             int[] array1 = { 1, 2, 3, 4, 5, 6, 7 };
  7             MoveFirstPartOfArrayToTheEnd(array1, 3);
  8             printArray(array1);
  9 
 10             //2
 11             int[] array2 = { 1, 2, 8, 4, 5, 6, 7 };
 12             MoveFirstPartOfArrayToTheEndValue(array2, 8);
 13             printArray(array2);
 14 
 15             //3
 16             int[] array3 = { 1, 2, 3, 4, 5, 6, 7, 8 };
 17             MoveSomeElementsToTheEnd(array3, 3, 3);
 18             printArray(array3);
 19 
 20             //4
 21             int[] array4 = { 1, 3, 3, 2, 4, 4, 4, 5 };
 22             removeDulplicatedElements(array4);
 23             printArray(array4);
 24 
 25         }
 26 
 27         private static void printArray(int[] array)
 28         {
 29             for (int i = 0; i < array.Length; i++)
 30             {
 31                 Console.Write(array[i]);                
 32             }
 33             Console.WriteLine();
 34         }
 35 
 36         public static void MoveFirstPartOfArrayToTheEnd(int[] array, int index)
 37         {
 38             if (index >= array.Length || index <= 0)
 39             {
 40                 throw new Exception("index must be greater than 0 and less than " + array.Length);
 41             }
 42 
 43             //Move the first part of array to a temp array
 44             int[] temp = new int[index];
 45             for (int i = 0; i < index; i++)
 46             {
 47                 temp[i] = array[i];
 48             }
 49 
 50             //Move forward the other element
 51             for (int i = 0; i < array.Length - index; i++)
 52             {
 53                 array[i] = array[i + index];
 54             }
 55 
 56             //Move the first part back to the end of array
 57             int j = 0;
 58             for (int i = array.Length - index; i < array.Length; i++)
 59             {
 60                 array[i] = temp[j];
 61                 j++;
 62             }
 63         }
 64 
 65         public static void MoveFirstPartOfArrayToTheEndValue(int[] array, int value)
 66         {
 67             bool move = false;
 68 
 69             //Search the value in the array
 70             for (int i = 0; i < array.Length; i++)
 71             {
 72                 if (array[i] == value)
 73                 {
 74 
 75                     //Move the first part if we find the value
 76                     MoveFirstPartOfArrayToTheEnd(array, i + 1);
 77                     move = true;
 78                     break;
 79                 }
 80             }
 81 
 82             if (!move)
 83             {
 84                 throw new Exception("No matched value is found in the array");
 85             }
 86         }
 87 
 88         public static void MoveSomeElementsToTheEnd(int[] array, int startIndex, int length)
 89         {
 90             if (startIndex < 0 || startIndex >= array.Length - 1)
 91             {
 92                 throw new Exception("startIndex must be greater than 0 and less than " + (array.Length - 1).ToString());
 93             }
 94 
 95             if (startIndex + length + 1 > array.Length)
 96             {
 97                 throw new Exception("Please provide a valid length");
 98             }
 99 
100             int[] temp = new int[length];
101             for (int i = 0; i < temp.Length; i++)
102             {
103                 temp[i] = array[startIndex + i];
104             }
105 
106             //Move forward the other element
107             for (int i = startIndex; i < array.Length - length; i++)
108             {
109                 array[i] = array[i + length];
110             }
111 
112             //Move the first part back to the end of array
113             int k = 0;
114             for (int i = array.Length - length; i < array.Length; i++)
115             {
116                 array[i] = temp[k];
117                 k++;
118             }
119         }
120 
121         public static void removeDulplicatedElements(int[] array)
122         {
123             List<int> temp = new List<int>();
124             for (int i = 0; i < array.Length; i++)
125             {
126                 if (array[i] == 0)
127                     continue;
128                 temp.Add(array[i]);
129                 for (int j = i + 1; j < array.Length; j++)
130                 {
131                     if (array[i] == array[j])
132                     {
133                         array[j] = 0;
134                     }
135                 }
136             }
137 
138             for (int i = 0; i < array.Length; i++)
139             {
140                 if (i < temp.Count)
141                 {
142                     array[i] = temp[i];
143                 }
144                 else
145                 {
146                     array[i] = 0;
147                 }
148             }
149         }
150     }
View Code

 

现在要求优化算法，空间复杂度为O（1），该如何去优化呢？

  1. 我们先看{1，2，3，4，5，6，7}3如何变成{4，5，6，7，1，2，3}的，首先前后部分各自反转变成{3，2，1，7，6，5，4}然后再把整个数组反转就变成了{4，5，6，7，1，2，3}。
  2. 同上，只要找到对应值的下标即可。
  3. 我们再看{1,2,3,4,5,6,7,8}  3  3  => {1,2,3,7,8,4,5,6}过程，前面不动的部分{1,2,3}先不管，且看后面{4,5,6,7,8}=>{7,8,4,5,6},同上面，各自反转尔后整体反转{6，5，4，8，7}=>{7,8,4,5,6}
  4. {1,3,3,2,4,4,4,5} => {1,3,2,4,5,0,0,0}的过程，先把重复的元素变成0，再依次移动到最后，{1,3,3,2,4,4,4,5} => {1,3,0,2,4,0,0,5}=>{1,3,2,4,0,0,5,0}=>{1,3,2,4,0,5,0,0}=>{1,3,2,4,5,0,0,0}

从上面可以看出都用到了一个通用的方法就是反转数组ReversArray().这个操作只需要申请一个临时变量。即空间复杂度为O(1)

  1 using System;
  2 using System.Collections.Generic;
  3 using System.Linq;
  4 using System.Text;
  5 using System.Threading.Tasks;
  6 
  7 namespace TransposeArray
  8 {
  9     class Program
 10     {
 11         static void Main(string[] args)
 12         {
 13             //1 {1,2,3,4,5,6,7} 3  => {4,5,6,7,1,2,3}
 14             int[] Array1 = new int[] { 1, 2, 3, 4, 5, 6, 7 };
 15             TransposeByIndex(Array1, 3);
 16             PrintArray(Array1);
 17             //2 {1,2,8,4,5,6,7} 8  => {4,5,6,7,1,2,8}
 18             int[] Array2 = new int[] { 1, 2, 8, 4, 5, 6, 7 };
 19             TransposeByValue(Array2, 8);
 20             PrintArray(Array2);
 21             //3 {1,2,3,4,5,6,7,8}  3  3  => {1,2,3,7,8,4,5,6}
 22             int[] Array3 = new int[] { 1, 2, 3, 4, 5, 6, 7, 8 };
 23             TransposeBySegment(Array3, 3, 3);
 24             PrintArray(Array3);
 25             //4  {1,3,3,2,4,4,4,5} => {1,3,2,4,5,0,0,0}
 26             int[] Array4 = new int[] { 1, 3, 3, 2, 4, 4, 4, 5 };
 27             RemoveDuplicated(Array4);
 28             PrintArray(Array4);
 29         }
 30 
 31         //Print the array.
 32         public static void PrintArray(int[] array)
 33         {
 34             for (int i = 0; i < array.Length; i++)
 35             {
 36                 Console.Write(array[i] + ", ");
 37 
 38             }
 39             Console.ReadLine();
 40         }
 41 
 42         // Reverse an Array
 43         public static void ReverseArray(int[] array, int left, int right)
 44         {
 45 
 46             for (; left < right; ++left, --right)
 47             {
 48                 int tmp = array[left];
 49                 array[left] = array[right];
 50                 array[right] = tmp;
 51             }
 52         }
 53 
 54         // Reverse thrice respectively.
 55         public static void TransposeByIndex(int[] array, int index)
 56         {
 57             ReverseArray(array, 0, index);
 58             ReverseArray(array, index + 1, array.Length - 1);
 59             ReverseArray(array, 0, array.Length - 1);
 60         }
 61 
 62         //Transpose by value
 63         public static void TransposeByValue(int[] array, int value)
 64         {
 65             for (int i = 0; i < array.Length; i++)
 66                 if (array[i] == value)
 67                 {
 68                     TransposeByIndex(array, i);
 69                 }
 70         }
 71 
 72         //Tanspose by segment
 73         public static void TransposeBySegment(int[] array, int StartIndex, int Length)
 74         {
 75             int div = StartIndex + Length - 1;
 76             int end = array.Length - 1;
 77             ReverseArray(array, StartIndex, div);
 78             ReverseArray(array, div + 1, end);
 79             ReverseArray(array, StartIndex, end);
 80         }
 81 
 82         //Romove duplicated Elements 
 83         public static void RemoveDuplicated(int[] array)
 84         {
 85             for (int i = array.Length - 1; i > 0; i--)
 86             {
 87                 for (int j = 0; j < i; j++)
 88                 {
 89                     if (array[i] == array[j])
 90                     {
 91                         array[i] = 0;
 92                     }
 93                 }
 94 
 95             }
 96 
 97             for (int i = 0; i < array.Length - 1; i++)
 98             {
 99                 for (int j = array.Length - 1; j > i; j--)
100                 {
101                     if (array[i] == 0 && array[j] != 0)
102                     {
103                         ReverseArray(array, i, j);
104                         ReverseArray(array, i, j - 1);
105                     }
106                 }
107             }
108 
109 
110         }
111     }
112 }
View Code

第4个问题可以用其他方法实现，不去调用ReversArray方法。

找到重复元素置为0，然后搜索为0的值，再把后面的元素前移，最后一个元素置0。空间复杂度也为O（1）

 

 

 

 

posted @ 2014-03-19 15:24  AnnieBy  阅读(24408)  评论(1)  编辑  收藏  举报