算法导论(4)归并排序

#pragma once
#include<limits>

/*合并两个已经排序好的子序列
两个已经排序好的子序列为src[startIndex]-src[middleIndex];src[middleIndex+1]-src[endIndex]
*/
template<class T>
void Merge(T *src, int startIndex, int middleIndex, int endIndex)
{
	int n1 = middleIndex - startIndex + 1;
	int n2 = endIndex - middleIndex;

	//使L[n1],R[n2]成为新的数组
	T *L = new T[n1 + 1];
	T *R = new T[n2 + 1];
	for (int i = 0; i < n1; i++) {
		L[i] = src[startIndex + i];
	}
	for (int i = 0; i < n2; i++) {
		R[i] = src[middleIndex + i + 1];
	}

	//哨兵牌,设置为该类型的最大值
	L[n1] = numeric_limits<T>::max();
	R[n2] = numeric_limits<T>::max();

	int i = 0, j = 0;
	for (int k = startIndex; k <= endIndex; k++) {
		if (L[i] <= R[j]) {
			src[k] = L[i];
			i++;
		}else{
			src[k] = R[j];
			j++;
		}
	}

	//删除动态分配的数组
	delete[] L;
	delete[] R;
}
/*
归并排序
*/
template<class T>
void MergeSort(T *src, int startIndex, int endIndex)
{
	if (startIndex < endIndex) {
		//这一部分相当于分治
		int midIndex = (startIndex + endIndex) / 2;
		MergeSort(src, startIndex, midIndex);
		MergeSort(src, midIndex + 1, endIndex);

		//这一部分相当于合并
		Merge(src, startIndex, midIndex, endIndex);
	}
}
posted @ 2016-04-24 12:12  ql698214  阅读(178)  评论(0编辑  收藏  举报