Merge Sort
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递归版
算法思想
采用分治(Divide and Conquer)思想
将长度为n的序列分为两个长度为n/2的子序列,分别对子序列进行排序,然后将这两个内部有序的子序列进行归并
代码实现
int merge(vector<int>& nums,int left,int mid,int right){ int index1 = left; int index2 = mid+1; vector<int> temp; while(index1 <= mid && index2 <= right) { if(nums[index1] < nums[index2]) temp.push_back(nums[index1++]); else temp.push_back(nums[index2++]); } while(index1 <= mid) temp.push_back(nums[index++]); while(index2 <= right) temp.push_back(nums[index2++]); for(int num : temp) nums[left++] = num;}void mergeSort(vector<int>& nums,int left,int right){ if(left < right) { int mid = left + (right-left)/2; mergeSort(nums,left,mid); mergesort(nums,mid+1,right); merge(nums,left,mid,right); } }
迭代版
算法思想
由于只有一个元素的序列天然有序,所以将序列中相邻的两个元素排序,形成n/2个长度为2的内部有序的子序列,然后在继续归并成n/4个长度的4的内部有序子序列,直至整个序列有序
把 n 个元素看成 n 个长度为1的有序子数组;
进行第一趟归并,即将上述的n个子数组两两合并,得到 n/2向上取整个长度为 2 的有序子数组;
重复第2步,每次子数组的长度加倍,直到归并成一个长度为 n 的有序数组为止。
Note:在某趟归并中,设各子数组的长度为step,则归并前a[0…n-1]中共有n/step个有序的子数组:a[0…step-1], a[step…2*step-1], … , a[(n/step)*step … n-1],若n/step为奇数,那么对最后一个有序的子数组不做任何处理
代码实现
void merge(vector<int>& nums,int left,int mid,int right){ int index1 = left; int index2 = mid+1; vector<int> temp while(index1 <= mid && index2 <= right) { if(nums[index1] < nums[index2]) temp.push_back(nums[index1++]); else temp.push_back(nums[index2++]); } while(index1 <= mid) temp.push_back(nums[index1++]); while(index2 <= right) temp.push_back(nums[index2++]);}void mergeSort(vector<int>& nums,int left,int right){ int step = 1; int size = right - left + 1; while(step < size) { int start = left; while(start + 2*step <= right) { merge(nums,start,start+step-1,start+2*step-1); start += 2*step; } if(start+step <= right) merge(nums,start,start+step-1,right); step << 1; }}
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