Algorithm---n digits remove K-bit and find the decimal number

Remove K-bit to find minimum number

An n-bit number, minus the K-bit, how to make the Left (n-k) digits in the original sequence of the number of the smallest

For example

8314925 minus 4 numbers, leaving 125 the smallest, note that there is a pre-order requirement, if not the order of course is 123.

Solution Solutions

The greedy algorithm is guaranteed to have an optimal solution at each location visited.

Idea One

Analysis : To find a total of n bits, the smallest number of M-bits. Then the number of the M-bit, the highest bit should be in the highest position of the original number to find, or not as the first bit, such as (get 3 digits of the smallest, if the hundred in the 25 to find, will not be the Hundred number):

The same 10-digit number can only be found in the number of points in the middle of the current search, the whole process is shown as follows:

Attention

In the interval there are several minimum values, take the maximum distance, to ensure that the next number of large enough to find space.

Reference Code

#include <iostream>#include<cstdlib>#include<cassert>using namespacestd;int*Q;intFindminindex (intArr[],intBegintEnd//[]{    if(Beg >end)return-1; intMINV =Arr[beg]; intMinindex =Beg;  for(inti = Beg +1; I <= end; ++i) {if(Arr[i] <MINV) {MINV=Arr[i]; Minindex=i; }    }    returnMinindex;}intGetremain (intArr[],intSizeintk) {Assert (Size> k && k >=0); intRev =0, Revindex =-1;  for(inti = size-k; i < size; ++i) {Revindex= Findminindex (arr, Revindex +1, i); Rev= Rev *Ten+Arr[revindex]; }    returnRev;}intMain () {intArr[] = {3,1,6,4,8,5,7}; size_t size=sizeof(arr)/sizeof(int); intRemainnum;  for(intK = size-1; K >0; --k) {intRemainnum = Getremain (arr, size, size-k); cout<<"When k ="<< k <<", the remaining value is:"<< Remainnum <<Endl; }}

Results

Analysis

Time complexity O (KN)

Two ideas

Analysis: From the go after, each visit one, compare the number in front of the bit, if bigger than that bit, decisive kill, for example:

The same goes backwards, knowing that the number of kills is K or access to the last, the whole process diagram as. Of course, the traversal to the end has not killed the K elements, indicating that the rest is already ascending, so that in the number left before the n-k, integrated into an integer is the minimum value.

Reference Code

#include <iostream>#include<cassert>using namespacestd;intGetremain (int*arr,intSizeintk) {Assert (Size> K && k >0); intTMP = size-K; intCur =0, pre; intRev =0;  while(k! =0&& Cur <size) {Pre= cur-1;  while(Pre >=0)        {            if(Arr[pre] >=Arr[cur]) {                 for(inti = Pre; i < size; ++i) arr[i]= arr[i+1]; --cur; --K; --size; }            --Pre; }        ++cur; }     for(inti =0; I < TMP; ++i) {rev= Rev *Ten+Arr[i]; }    returnRev;}intMain () {intArr[] = {3,1,6,4,8,5,7}; size_t size=sizeof(arr)/sizeof(int); intRemainnum; Remainnum= Getremain (arr, size,3); cout<<"When k ="<<3<<", the remaining value is:"<< Remainnum <<Endl;}

Analysis

Time complexity O (KN)

Algorithm---n digits remove K-bit and find the decimal number