《算法导论》第3版2.3.1分治法里提出了归并排序的算法
归并排序算法完全遵循分治模式
1）分解：分解待排序的n个元素的序列成各具n/2个元素的两个子序列
2）解决：使用归并排序递归地排序两个子序列
3）合并：合并两个已排序的子序列以产生已排序的答案
设置哨兵，它不可能为较小值
void mergeWithSentry(int A[], int left, int mid, int right) { int n1 = mid - left + 1; int n2 = right - mid; int *L = new int[n1 + 1]; int *R = new int[n2 + 1]; for (int i = 0; i < n1; ++i) L[i] = A[left + i]; for (int j = 0; j < n2; ++j) R[j] = A[mid + 1 + j]; L[n1] = INT_MAX; R[n2] = INT_MAX; int i = 0, j = 0; int k = left; while (k <= right) { if (L[i] <= R[j]) A[k++] = L[i++]; else A[k++] = R[j++]; } delete L; L = NULL; delete R; R = NULL; } void merge_sort(int A[], int left, int right) { if (left < right) { int mid = left + (right - left) / 2; merge_sort(A, left, mid); merge_sort(A, mid + 1, right); // mergeWithSentry(A, left, mid, right); mergeWithoutSentry(A, left, mid, right); } }《算法导论》第3版习题2.3-2提出
重写过程merge，使之不使用哨兵，而是一旦数组L或R的所有元素均被复制回A就立刻停止，然后把另一个数组的剩余部分复制回A。
void mergeWithoutSentry(int A[], int left, int mid, int right) { int n1 = mid - left + 1; int n2 = right - mid; int *L = new int[n1]; int *R = new int[n2]; for (int i = 0; i < n1; ++i) L[i] = A[left + i]; for (int j = 0; j < n2; ++j) R[j] = A[mid + 1 + j]; int i = 0, j = 0; int k = left; while (i < n1 && j < n2) { if (L[i] <= R[j]) A[k++] = L[i++]; else A[k++] = R[j++]; } while (i < n1) A[k++] = L[i++]; while (j < n2) A[k++] = R[j++]; delete L; L = NULL; delete R; R = NULL; } void merge_sort(int A[], int left, int right) { if (left < right) { int mid = left + (right - left) / 2; merge_sort(A, left, mid); merge_sort(A, mid + 1, right); // mergeWithSentry(A, left, mid, right); mergeWithoutSentry(A, left, mid, right); } }注意不要对L和R数组的长度计算错误
不要遗漏对L和R数组的初始化，以及初始化时的下标
注意k的取值范围不要和i,j的取值范围混淆
记得不要遗漏i,j,k的自增、在什么时候自增
设置哨兵：注意不要遗漏哨兵的设置
不使用哨兵：注意L或R数组是否未遍历完，若有剩余元素注意不能遗漏
void merge(int A[], int left, int mid, int right, int temp[]) { int i = left, j = mid + 1; int k = left; while (i <= mid && j <= right) { if (A[i] <= A[j]) temp[k++] = A[i++]; else temp[k++] = A[j++]; } while(i <= mid) temp[k++] = A[i++]; while (j <= right) temp[k++] = A[j++]; // 注意 i <= right 不要写成 i < right for (int i = left; i <= right; ++i) A[i] = temp[i]; } void merge_sort(int A[], int left, int right, int temp[]) { if (left < right) { int mid = left + (right - left) / 2; merge_sort(A, left, mid, temp); merge_sort(A, mid + 1, right, temp); merge(A, left, mid, right, temp); } }完整源代码放于github
相关代码：
合并两个有序数组-LeetCode-088. Merge Sorted Array
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