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  • 常用的排序代码

    常用排序算法的代码:

      1 class Sort {
      2 public:
      3     //  冒泡排序
      4     int* bubbleSort(int* A, int n) {
      5         // write code here
      6         int i,j;
      7         for(i = n - 1; i >= 0; i--){
      8             for(j = 0; j < i; j++)
      9                 if(A[j] > A[j + 1])
     10                     swap(A[j],A[j + 1]);
     11         }
     12         return A;
     13     }
     14 public:
     15     //  选择排序
     16     int* selectionSort(int* A, int n) {
     17         // write code here
     18         int i,j,temp;
     19         for(i = 0; i < n; i++){
     20             temp = i;
     21             for(j = i + 1; j < n; j++)
     22                 if(A[j] < A[temp])
     23                     temp = j;
     24             swap(A[i],A[temp]);
     25         }
     26         return A;
     27     }
     28 public:
     29     //  插入排序
     30     int* insertionSort(int* A, int n) {
     31         // write code here
     32         int i,j,temp;
     33         for(i = 1; i < n; i++){
     34             temp = A[i];
     35             for(j = i; j > 0 && temp < A[j - 1]; j--)
     36                 A[j] = A[j - 1];
     37             A[j] = temp;
     38         }
     39         return A;
     40     }
     41 public:
     42     //  归并排序
     43     int* mergeSort(int* A, int n) {
     44         // write code here
     45         if(n <= 0 || A == NULL)
     46             return A;
     47         Msort(A,0,n-1);
     48         return A;
     49     }
     50     void Msort(int *A,int left,int right){
     51         if(left == right)
     52             return ;
     53         else{
     54             int mid = (left + right) / 2;
     55             Msort(A,left,mid);
     56             Msort(A,mid + 1,right);
     57             Merge(A,left,mid,right);
     58         }
     59     }
     60     void Merge(int *A,int left,int mid,int right){
     61         int leftStart = left,leftEnd = mid,rightStart = mid + 1,rightEnd = right;
     62         int i = 0;
     63         int *tempA = new int[right - left + 1];
     64         while(leftStart <= leftEnd && rightStart <= rightEnd){
     65             if(A[leftStart] <= A[rightStart])
     66                 tempA[i++] = A[leftStart++];
     67             else
     68                 tempA[i++] = A[rightStart++];
     69         }
     70         while(leftStart <= leftEnd)
     71             tempA[i++] = A[leftStart++];
     72         while(rightStart <= rightEnd)
     73             tempA[i++] = A[rightStart++];
     74         for(int j = left,i = 0; j <= right; i++) //注意把tempA中元素拷贝到A中,下标从left到right
     75             A[j++] = tempA[i];
     76         delete []tempA;
     77         tempA = NULL; //释放空间后将指针置空,防止产生野指针     
     78     }
     79 public:
     80     //  快速排序
     81     int* quickSort(int* A, int n) {
     82         // write code here
     83         if(A == NULL || n <= 0)
     84             return A;
     85         else
     86             Q_sort(A,0,n - 1);
     87         return A;
     88     }
     89     void Q_sort(int *A,int left,int right){
     90         if(left >= right)
     91             return;
     92         int mid = (left + right) / 2;
     93         int i,last = left;
     94         swap(A[left],A[mid]);
     95         for(i = left + 1; i <= right; i++){
     96             if(A[i] < A[left])
     97                 swap(A[i],A[++last]);
     98         }
     99         swap(A[left],A[last]);
    100         Q_sort(A,left,last - 1);
    101         Q_sort(A,last + 1,right);
    102     }
    103 public:
    104     //  堆排序
    105     int* heapSort(int* A, int n) {
    106         // write code here
    107         int i;
    108         for(i = n / 2 - 1; i >= 0; i--)
    109             PercDown(A,i,n);
    110         for(i = n - 1; i > 0; i--){
    111             swap(A[0],A[i]);
    112             PercDown(A,0,i);
    113         }
    114         return A;
    115     }
    116     void PercDown(int *A,int p,int n){
    117         int parent,child,temp = A[p];
    118         for(parent = p; parent * 2 + 1 < n; parent = child){
    119             child = parent * 2 + 1;
    120             if(child + 1 != n && A[child] < A[child + 1])
    121                 child++;
    122             if(A[child] < temp)  //注意是建立大顶堆
    123                 break;
    124             else
    125                 A[parent] = A[child];
    126         }
    127         A[parent] = temp;    
    128     }
    129 public:
    130     //  希尔排序
    131     int* shellSort(int* A, int n) {
    132         // write code here
    133         if(A == NULL || n <= 0)
    134             return A;
    135         int step,i,j,temp;
    136         for(step = n / 2; step > 0; step /= 2){
    137             for(i = step; i < n; i++){
    138                 temp = A[i];
    139                 for(j = i; j >= step && temp < A[j - step]; j -= step)
    140                     A[j] = A[j - step];
    141                 A[j] = temp;
    142             }
    143         }
    144         return A;
    145     }
    146 public:
    147     //  计数排序
    148     int* countingSort(int* A, int n) {
    149         // write code here
    150         if(A == NULL || n <= 0)
    151             return A;
    152         int max = 0;
    153         int min = A[0];
    154         for(int i = 0; i < n; i++){  //求大值,代表桶的最大数目
    155             if(A[i] > max)
    156                 max = A[i];
    157             if(A[i] < min)
    158                 min = A[i];
    159         }
    160         int *tempA = new int[max + 1]();  //new申请内存空间,并初始化为0
    161         for(int i = 0; i < n; i++)
    162             tempA[A[i]]++;
    163         int index = 0;
    164         for(int i =min,j; i <= max; i++){
    165             for(j = 0; j < tempA[i]; j++)
    166                 A[index++] = i;
    167         }
    168         delete []tempA;
    169         tempA = NULL;
    170         return A;        
    171     }
    172 public:
    173     //  基数排序
    174     int* radixSort(int* A, int n) {
    175         // write code here
    176         if(A== NULL || n <= 1)
    177             return A;
    178         vector<vector<int>> res(10); //创建10个桶
    179         int max = 0;
    180         for(int i = 0; i < n; i++)
    181             if(A[i] > max)
    182                 max = A[i];
    183         int count = 0;
    184         while(max){
    185             count++;
    186             max /= 10;
    187         }
    188         
    189         for(int i = 0; i < count; i++){
    190             for(int j = 0; j < n; j++){
    191                 int num = A[j] % (int)pow(10,i + 1) / (int)pow(10,i);  //num代表个、十、百、千位上的数字
    192                 res[num].push_back(A[j]);
    193             }
    194             int index = 0;
    195             for(auto iter1 = res.begin(); iter1 != res.end(); iter1++){
    196                 for(auto iter2 = iter1->begin(); iter2 != iter1->end(); iter2++)
    197                     A[index++] = *iter2;
    198             }
    199             for(int i = 0; i < 10; i++)
    200                 res[i].clear();
    201         }
    202         return A;
    203     }
    204 };
    只为训练自己,时刻锤炼一个程序员最基本的技能!
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  • 原文地址:https://www.cnblogs.com/coding-wtf/p/6891044.html
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