三大排序算法
merge sort
空间复杂度:O(n)(sort list是O(1))
时间复杂度为O(nlgn)
稳定性:稳定排序
先局部有序,再整体有序
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 public class Solution { 2 /** 3 * @param A an integer array 4 * @return void 5 */ 6 public void sortIntegers(int[] A) { 7 int[] temp = new int[A.length]; 8 mergeSort(A, 0, A.length - 1, temp); 9 } 10 11 public void mergeSort(int[] A, int start, int end, int[] temp) { 12 if (start >= end) { 13 return; 14 } 15 mergeSort(A, start, (start + end) / 2, temp); 16 mergeSort(A, (start + end) / 2 + 1, end, temp); 17 18 merge(A, start, end, temp); 19 } 20 21 public void merge(int[] A, int start, int end, int[] temp) { 22 int mid = (start + end) / 2; 23 int left = start; 24 int right = mid + 1; 25 int index = start; 26 27 while (left <= mid && right <= end) { 28 if (A[left] <= A[right]) { 29 temp[index++] = A[left++]; 30 } else { 31 temp[index++] = A[right++]; 32 } 33 } 34 while (left <= mid) { 35 temp[index++] = A[left++]; 36 } 37 while (right <= end) { 38 temp[index++] = A[right++]; 39 } 40 for (int i = start; i <= end; i++) { 41 A[i] = temp[i]; 42 } 43 } 44 }
quick sort
空间复杂度:O(1)的额外空间
时间复杂度:期望时间复杂度为O(nlgn),最坏情况下时间复杂度为O(n^2)
稳定性:不稳定排序
先整体有序,再局部有序
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 public class Solution { 2 /** 3 * @param A an integer array 4 * @return void 5 */ 6 public void sortIntegers(int[] A) { 7 if (A == null || A.length == 0) { 8 return; 9 } 10 quickSort(A, 0, A.length - 1); 11 } 12 13 public void quickSort(int[] A, int start, int end) { 14 if (start >= end) { 15 return; 16 } 17 int left = start; 18 int right = end; 19 int pivot = A[(start + end) / 2]; 20 21 while (left <= right) { 22 while (left <= right && A[left] < pivot) { 23 left++; 24 } 25 while (left <= right && A[right] > pivot) { 26 right--; 27 } 28 if (left <= right) { 29 int tmp = A[left]; 30 A[left] = A[right]; 31 A[right] = tmp; 32 left++; 33 right--; 34 } 35 } 36 37 quickSort(A, start, right); 38 quickSort(A, left, end); 39 } 40 }
heap sort
空间复杂度:O(1)
时间复杂度: O(nlgn)
步骤:把数组A建立成一个大根堆(从左到右shiftup操作)。然后把堆顶元素(即最大值)与堆末尾元素交换,再shiftdown,重复n次。
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
public class Solution { /** * @param A: an integer array * @return: nothing */ public void sortIntegers2(int[] A) { if (A == null) { return; } buildHeap(A); int size = A.length; for (int i = A.length - 1; i >= 0; i--) { swap(A, 0, i); size--; shiftDown(A, 0, size); } } public void buildHeap(int[] A) { for (int i = 0; i < A.length; i++) { shiftUp(A, i); } } public void shiftUp(int[] A, int cur) { if (cur == 0) { return; } int father = cur % 2 == 0 ? cur / 2 - 1 : cur / 2; if (A[father] < A[cur]) { swap(A, father, cur); } shiftUp(A, father); } public void shiftDown(int[] A, int cur, int size) { if (cur >= size - 1) { return; } int left = cur * 2 + 1; int right = cur * 2 + 2; int max_child = -1; if (left <= size - 1 && right <= size - 1) { max_child = A[left] < A[right] ? right : left; } else if (left <= size - 1) { max_child = left; } else { return; } if(A[cur] < A[max_child]) { swap(A, cur, max_child); shiftDown(A, max_child, size); } } public void swap(int[] A, int i, int j) { int tmp = A[i]; A[i] = A[j]; A[j] = tmp; } }