归并排序(merge sort)
2016-09-07 13:24
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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.
如 设有数列{6,202,100,301,38,8,1}
初始状态:6,202,100,301,38,8,1
第一次归并后:{6,202},{100,301},{8,38},{1},比较次数:3;
第二次归并后:{6,100,202,301},{1,8,38},比较次数:4;
第三次归并后:{1,6,8,38,100,202,301},比较次数:4;
总的比较次数为:3+4+4=11,;
逆序数为14;
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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.
归并操作
归并操作(merge),也叫归并算法,指的是将两个顺序序列合并成一个顺序序列的方法。如 设有数列{6,202,100,301,38,8,1}
初始状态:6,202,100,301,38,8,1
第一次归并后:{6,202},{100,301},{8,38},{1},比较次数:3;
第二次归并后:{6,100,202,301},{1,8,38},比较次数:4;
第三次归并后:{1,6,8,38,100,202,301},比较次数:4;
总的比较次数为:3+4+4=11,;
逆序数为14;
/* C program for Merge Sort */ #include<stdlib.h> #include<stdio.h> // Merges two subarrays of arr[]. // First subarray is arr[l..m] // Second subarray is arr[m+1..r] void merge(int arr[], int l, int m, int r) { int i, j, k; int n1 = m - l + 1; int n2 = r - m; /* create temp arrays */ int L[n1], R[n2]; /* Copy data to temp arrays L[] and R[] */ for (i = 0; i < n1; i++) L[i] = arr[l + i]; for (j = 0; j < n2; j++) R[j] = arr[m + 1+ j]; /* Merge the temp arrays back into arr[l..r]*/ i = 0; // Initial index of first subarray j = 0; // Initial index of second subarray k = l; // Initial index of merged subarray while (i < n1 && j < n2) { if (L[i] <= R[j]) { arr[k] = L[i]; i++; } else { arr[k] = R[j]; j++; } k++; } /* Copy the remaining elements of L[], if there are any */ while (i < n1) { arr[k] = L[i]; i++; k++; } /* Copy the remaining elements of R[], if there are any */ while (j < n2) { arr[k] = R[j]; j++; k++; } } /* l is for left index and r is right index of the sub-array of arr to be sorted */ void mergeSort(int arr[], int l, int r) { if (l < r) { // Same as (l+r)/2, but avoids overflow for // large l and h int m = l+(r-l)/2; // Sort first and second halves mergeSort(arr, l, m); mergeSort(arr, m+1, r); merge(arr, l, m, r); } } /* UTILITY FUNCTIONS */ /* Function to print an array */ void printArray(int A[], int size) { int i; for (i=0; i < size; i++) printf("%d ", A[i]); printf("\n"); } /* Driver program to test above functions */ int main() { int arr[] = {12, 11, 13, 5, 6, 7}; int arr_size = sizeof(arr)/sizeof(arr[0]); printf("Given array is \n"); printArray(arr, arr_size); mergeSort(arr, 0, arr_size - 1); printf("\nSorted array is \n"); printArray(arr, arr_size); return 0; }
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