Array al [0, mid-1] and Al [mid, num-1] are their respective orders, the two sub-ordered segments of array al [0, num-1] merge, obtain the overall order of Al [0, num-1 ].

Problem description:

Array al [0, mid-1] and Al [mid, num-1] are their respective orders, the two sub-ordered segments of array al [0, num-1] merge, obtain the overall order of Al [0, num-1. The space complexity is O (1 ). Note: The al [I] element supports the '<' operator.

Idea: because the space complexity is required to be O (1), Merge Sorting 1 cannot be used to traverse 0 ~ Mid-1, compare it with a [Mid], if a [Mid] <A [I]; then switch it to 22, traverse a [Mid ~ cyclically ~ Length]. If a [Mid]> A [Mid + 1] After switching in 1, the system performs switching, performs insertion sorting, and inserts a [Mid] to the correct position.
Program

# Include <iostream>
Using namespace STD;
Void insertsort (int A [], int index, int length) {int temp; For (INT I = index; I <length; I ++) {if (a [I]> = A [I + 1]) {temp = A [I]; A [I] = A [I + 1]; A [I + 1] = temp ;}}}
Void sort (int A [], int mid, int length) {int temp; For (INT I = 0; I <= mid-1; I ++) {if (a [Mid] <A [I]) {temp = A [I]; A [I] = A [Mid]; A [Mid] = temp; insertsort (A, mid, length-1 );}}}
Int main () {int A [11] = {1, 4, 6, 7, 10, 2, 3, 8, 9, 15, 16}; sort (, 5, 11); For (INT I = 0; I <11; I ++) cout <A [I] <""; return 0 ;}