剖析递归行为和递归行为时间复杂度的估算
一个递归行为的例子
T(N) = a*T(N/b) + O(N^d)
1) log(b,a) > d -> 复杂度为O(N^log(b,a))
2) log(b,a) = d -> 复杂度为O(N^d * logN)
3) log(b,a) < d -> 复杂度为O(N^d)
例:
归并排序
1 public static void mergeSort(int[] arr) { 2 if (arr == null || arr.length < 2) { 3 return; 4 } 5 mergeSort(arr, 0, arr.length - 1); 6 } 7 8 public static void mergeSort(int[] arr, int l, int r) { 9 if (l == r) { 10 return; 11 } 12 int mid = l + ((r - l) >> 1); 13 mergeSort(arr, l, mid); 14 mergeSort(arr, mid + 1, r); 15 merge(arr, l, mid, r); 16 } 17 18 public static void merge(int[] arr, int l, int m, int r) { 19 int[] help = new int[r - l + 1]; 20 int i = 0; 21 int p1 = l; 22 int p2 = m + 1; 23 while (p1 <= m && p2 <= r) { 24 help[i++] = arr[p1] < arr[p2] ? arr[p1++] : arr[p2++]; 25 } 26 while (p1 <= m) { 27 help[i++] = arr[p1++]; 28 } 29 while (p2 <= r) { 30 help[i++] = arr[p2++]; 31 } 32 for (i = 0; i < help.length; i++) { 33 arr[l + i] = help[i]; 34 } 35 }
T(N) = 2 * T(N/2) + O(N);
log(b,a) = 1==d
所以 T(N) = O(nlogn)