Problem illustration:
given a n*n matrix, print its transition, for example , 90 degree clockwise,using only constant additional space
analysis:
using O(n^2) space is common,but for constant assistant space, we need to adopt constant space
solution:
use in place replacement,
take
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 17 18 19 20
for example
the expected output is
21 16 11 6 1
22 17 12 7 2
23 18 13 8 3
24
19 14 9 4
25 20 15 10 5
after transition,value (0,0) is expected to be transferred to pos(0,4),while(0,4)→(4,4),(4,4) to (4,0),(4,0)to (0,0),which is the same with our
so this four elements are called a cycle(for more clear explanation, please refer to situ permutation)
two keypoints:
1.with the same example,since we use (0,0) to cover (4,0),then the original value at (4,0) must be stored before coverage or the whole cycle will be initialized
to the cycle header
2.edge condition handling
source code for above example:
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#include<iostream>#include<cstdio>#include<string.h>#include<cstring>#include<string>#include<algorithm>using
namespace std;const
int maxn = 100;int
matrix1[maxn][maxn];int
matrix2[maxn][maxn];void
trans(int
h,int w){ for(int
i = 0; i < h; ++i) { for(int
j = 0; j < w; ++j) { //要旋转的元素是m1[i+1][j+1]，转过去的位置是m2[j][n-i] matrix2[j][w-i-1] = matrix1[i][j]; } } return;}inline
void getPos(int
orix,int
oriy,int
*newx,int
*newy){ (*newx) = oriy; (*newy) = 4 - orix;// return 0;}void
inPlaceTrans(int
i,int j){ //(i,j)为转换的起点 int
total = 25; int
cnt = 0; int
x = -1,y = -1; int
orix = i,oriy = j; bool
visit[5][5]; memset(visit,0,sizeof(visit)); visit[i][j] = 1; int
tmp1,tmp2; while(cnt < total){ //我去，移反了，所有的值都initialize成圈头值了 //a new cycle orix = i; oriy = j; getPos(i,j,&x,&y); //(i,j)的目标地址 visit[i][j] = true;//这个位置的元素已经被移到目标地址 tmp1 = matrix1[i][j]; while(!(x == orix && y == oriy)){ //这个圈没有到头 tmp2 = matrix1[x][y]; matrix1[x][y] = tmp1; ++cnt; //移过去了一个元素 i = x; j = y; visit[i][j] = true; tmp1 = tmp2; getPos(i,j,&x,&y); } matrix1[x][y] = tmp1; bool
get = false; for(int
u = 0; u < 5; ++u){ for(int
v = 0; v < 5; ++v){ if(visit[u][v] == 0){ //这个元素还没有被移到目标地址 i = u; j = v; get = true; break; } } if(get) break; } if(!get){ //所有元素都被移到目标地址 break; } }}int
main(){ freopen("1.txt","r",stdin); freopen("2.txt","w",stdout); cout << "helloworld\n"; for(int
i = 0; i < 5; ++i) { for(int
j = 0; j < 5; ++j) { scanf("%d",matrix1[i]+j); } } inPlaceTrans(0,0); for(int
i = 0; i < 5; ++i) { for(int
j = 0; j < 5; ++j) { printf("%d ",matrix1[i][j]); } cout << endl; } cout << "hello world\n"; return
0;} |
InPlace Transition of a matrix,布布扣,bubuko.com
InPlace Transition of a matrix
原文：http://www.cnblogs.com/warmfrog/p/3695011.html