②进行两两归q记录关键字有序，得到 n/2 个长度ؓ 2 的有序子表；
③重复第②步直到所有记录归q成一个长度ؓ n 的有序表为止?br /> ?归ƈ排序法实例
对于归ƈ排序法q类的分ȝ法，其核心就?分解"?递归求解"。对?分解"实例Q会在下面分析msort()Ҏ中给出。我们先看合q的q程?br /> 以下面描q的序列ZQ在索引范围内[first , last)的序列还有九个整数元素，它由索引范围为[first , mid]的四个元素有序子列表A和烦引范围[mid , last]的五个元素有序子列表Bl成?br />
步骤1Q比较arr[indexA]=7与arr[indexB]=12。将较小的元?复制到数ltempArr的烦引indexC处。ƈindexA和indexC都指向下一个位|?br />
步骤2Q比较arr[indexA]=10与arr[indexB]=12。将较小的元?0复制到数ltempArr的烦引indexC处。ƈindexA和indexC都指向下一个位|?br />
步骤3Q比较arr[indexA]=19与arr[indexB]=12。将较小的元?2复制到数ltempArr的烦引indexC处。ƈindexB和indexC都指向下一个位|?br />
步骤4-7Q依ơ成Ҏ较两子表的元素将17Q?9Q?1Q?5复制到数ltempArr。此ӞindexA到达子表A的未?indexA = mid)QindexB引用的gؓ30?br />
步骤8-9Q将未到N的子表剩余数据复制到tempArr中?br />
步骤10Q将tempArr复制到原始数据arr中?br />
?归ƈ排序法的实?br /> 了解了合q过E，那么理解下面的代码ƈ不是一仉事。下面提供了归ƈ法的非泛型版本和泛型版本?br />
public class MergeSort {
public static void sort(Object[] arr) {
//create a temporary array to store partitioned elements Object[] tempArr = arr.clone(); //call msort with arrays arr and tempArr along //with the index range msort(arr, tempArr, 0, arr.length);
} public static <T extends Comparable<? super T>> void sort(T[] arr) { //create a temporary aray to store partitioned elements T[] tempArr = (T[]) arr.clone(); //call msort with arrays arr and tempArr along //with the index range msort(arr, tempArr, 0, arr.length); } private static void msort(Object[] arr, Object[] tempArr, int first, int last) { //if the sublist has more than 1 element continue if (first + 1 < last) { //for sublists of size 2 or more, call msort() //for the left and right sublists and than //merge the sorted sublists using merge() int midpt = (last + first) / 2; msort(arr, tempArr, first, midpt); msort(arr, tempArr, midpt, last); //if list is already sorted, just copy src to //dest; this is an optimization that results in faster //sorts for nearly ordered lists if (((Comparable) arr[midpt - 1]).compareTo(arr[midpt]) <= 0) return; //the elements in the ranges [first,mid] and //[mid,last] are ordered;merge the ordered sublists //into an ordered sequence in the range [first , last] //using the temporary array int indexA, indexB, indexC; //set indexA to scan sublist A with rang [first , mid] //and indexB to scan sublist B with rang [mid , last] indexA = first; indexB = midpt; indexC = first; //while both sublists are not exhausted, compare //arr[indexA] and arr[indexB]; copy the smaller //to tempArr while (indexA < midpt && indexB < last) { if (((Comparable) arr[indexA]).compareTo(arr[indexB]) < 0) { tempArr[indexC] = arr[indexA]; //copyto tempArr indexA++; //increment indexA } else { tempArr[indexC] = arr[indexB]; //copyto tempArr indexB++; //increment indexB } indexC++; //increment indexC } //copy the tail of the sublist that is not exhausted while (indexA < midpt) { tempArr[indexC++] = arr[indexA++]; //copy to tempArr } while (indexB < last) { tempArr[indexC++] = arr[indexB++]; //copy to tempArr } //copy elements form temporary array to original array for (int i = first; i < last; i++) arr[i] = tempArr[i]; } }
}上述代码中最核心的msort()Ҏ是一递归法。下图说明了msort()Ҏ中子列表的分割与合ƈ?nbsp;
?归ƈ排序法的效?br /> 归ƈ排序的最坏情况与q_情况q行旉都ؓO(nlog2n)。假定数l具有n=2k个元素。如下图Q?br />
在层?上对msort()Ҏ的第一个调用会产生两个递归调用Q这两个递归调用产生长度为n/2的两个半部分列表Q而merge()Ҏ上qC个半部分列表l合的一个有序的n元素列表Q在层数1上存在两个msort()Ҏ的调用，每个调用又会产生另外两个寚w度ؓn/4的列表的递归调用。每个合q会两个长度ؓn/4的子列表q接Z个长度ؓn/2的有序列表；在层?上存在对merge()Ҏ?=22个调用，每个调用会创Z个长度ؓn/4的有序列表。通常Q在层数i上存在对merge()Ҏ?i个调用，每个调用会创Z个长度ؓn/2i的有序子列表?br /> 层数0Q存在对merge()Ҏ?=20ơ调用。这个调用对n个元素排序?br /> 层数1Q存在对merge()Ҏ?=21ơ调用。这个调用对n/2个元素排序?br /> 层数2Q存在对merge()Ҏ?=22ơ调用。这个调用对n/4个元素排序?br /> ......
层数iQ存在对merge()Ҏ?iơ调用。这个调用对n/i个元素排序?br /> 在树中的每一层，合ƈ涉及hU性运行时间的n/2i个元素，q个U性运行时间需要少于n/2iơ的比较。在层数i上组合的2i个合q操作需要少?i*n/2i=nơ的比较。假定n=2k,分割q程会在n/2k=1的k层数上终止。那么所有层上完成的工作总量?k*n = nlog2n。因此msort()Ҏ的最坏情冉|率ؓO(nlog2n)?
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假定q个数组的序是排好的Q然后从头往后，如果有数比当前外层元素的值大Q则这个数的位|往后挪Q直到当前外层元素的值大于或者等于它前面的位|ؓ止?br /> ?插入排序法实例
用五个名?Monroe,Chin,Flores,Stein和Dare)的列表的插入排序法ZQ?br /> Monroe 从Monroe开?br />
处理名字Chin Chine Monroe Chin插入C|?QMonroeUd至位|?
处理名字Flores Chine Flores Monroe Flores插入C|?QMonroeUd至位|?
处理名字Stein Chine Flores Monroe Stein Stein位置正确
处理名字Dare Chine Dare Flores Monroe Stein Dare插入在位|?Q列表尾部向右移?nbsp;
?插入排序法的实?br />
public class InsertSort { //sort an array of elements using insertion sort public static <T extends Comparable<? super T>> void sort(T[] arr) { int i, j, n = arr.length; T target; /** * place element at index i into the sublist * from index 0 to i-1 where 1<= i, * so it is in the correct positon */ for (i = 1; i < n; i++) { //index j scans down list from index i looking for //correct position to locate target; assigns it to //arr at index j j = i; target = arr[i]; //locate insertion point by scanning downward as long //as target < arr[j] and we have not encountered the //beginning of the array while (j > 0 && target.compareTo(arr[j - 1]) < 0) { //shift elements up list to make room for insertion arr[j] = arr[j - 1]; j--; } //the location is found;insert target arr[j] = target; } }}?插入排序法的效?br /> 假定n是数l的长度Q那么插入排序需要n-1遍。对于通用的遍i来说Q插入操作从arr[0]到arr[i-1]的子列表中，q且需要^均i/2ơ比较。比较的q_L为：
T(n) = 1/2 + 2/2 + 3/2 + ...... + (n-2)/2 + (n-1)/2 = n(n-1)/4
ҎT(n)的主，插入排序法的^均运行时间ؓO(n2)。最好情况ؓO(n)Q最坏情况ؓO(n2)?
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