标签：
M erge sort is based on the divide-and-conquer paradigm. Its worst-case running time has a lower order of growth
than insertion sort. Since we are dealing with subproblems, we state each subproblem as sorting a subarray A[p .. r].
Initially, p = 1 and r = n, but these values change as we recurse through subproblems.
To sort A[p .. r]:
1. Divide Step
If a given array A has zero or one element, simply return; it is already sorted. Otherwise, split A[p .. r]
into two subarrays A[p .. q] and A[q + 1 .. r], each containing about half of the elements of A[p .. r].
That is, q is the halfway point of A[p .. r].
2. Conquer Step
Conquer by recursively sorting the two subarrays A[p .. q] and A[q + 1 .. r].
3. Combine Step
Combine the elements back in A[p .. r] by merging the two sorted subarrays A[p .. q] and A[q + 1 .. r] into
a sorted sequence. To accomplish this step, we will define a procedure MERGE (A, p, q, r).
Note that the recursion bottoms out when the subarray has just one element, so that it is trivially sorted.
1 /* C program for Merge Sort */ 2 #include<stdlib.h> 3 #include<stdio.h> 4 5 // Merges two subarrays of arr[]. 6 // First subarray is arr[l..m] 7 // Second subarray is arr[m+1..r] 8 void merge(int arr[], int l, int m, int r) 9 { 10 int i, j, k; 11 int n1 = m - l + 1; 12 int n2 = r - m; 13 14 /* create temp arrays */ 15 int L[n1], R[n2]; 16 17 /* Copy data to temp arrays L[] and R[] */ 18 for (i = 0; i < n1; i++) 19 L[i] = arr[l + i]; 20 for (j = 0; j < n2; j++) 21 R[j] = arr[m + 1+ j]; 22 23 /* Merge the temp arrays back into arr[l..r]*/ 24 i = 0; // Initial index of first subarray 25 j = 0; // Initial index of second subarray 26 k = l; // Initial index of merged subarray 27 while (i < n1 && j < n2) 28 { 29 if (L[i] <= R[j]) 30 { 31 arr[k] = L[i]; 32 i++; 33 } 34 else 35 { 36 arr[k] = R[j]; 37 j++; 38 } 39 k++; 40 } 41 42 /* Copy the remaining elements of L[], if there 43 are any */ 44 while (i < n1) 45 { 46 arr[k] = L[i]; 47 i++; 48 k++; 49 } 50 51 /* Copy the remaining elements of R[], if there 52 are any */ 53 while (j < n2) 54 { 55 arr[k] = R[j]; 56 j++; 57 k++; 58 } 59 } 60 61 /* l is for left index and r is right index of the 62 sub-array of arr to be sorted */ 63 void mergeSort(int arr[], int l, int r) 64 { 65 if (l < r) 66 { 67 // Same as (l+r)/2, but avoids overflow for 68 // large l and h 69 int m = l+(r-l)/2; 70 71 // Sort first and second halves 72 mergeSort(arr, l, m); 73 mergeSort(arr, m+1, r); 74 75 merge(arr, l, m, r); 76 } 77 } 78 79 /* UTILITY FUNCTIONS */ 80 /* Function to print an array */ 81 void printArray(int A[], int size) 82 { 83 int i; 84 for (i=0; i < size; i++) 85 printf("%d ", A[i]); 86 printf("\n"); 87 } 88 89 /* Driver program to test above functions */ 90 int main() 91 { 92 int arr[] = {12, 11, 13, 5, 6, 7}; 93 int arr_size = sizeof(arr)/sizeof(arr[0]); 94 95 printf("Given array is \n"); 96 printArray(arr, arr_size); 97 98 mergeSort(arr, 0, arr_size - 1); 99 100 printf("\nSorted array is \n"); 101 printArray(arr, arr_size); 102 return 0; 103 }
标签：
原文地址：http://www.cnblogs.com/guxuanqing/p/5849106.html
