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 };
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