给定一组n个符号,大小k和来自符号集合的非重复字符的长度k的组合,仅写入迭代算法来打印可以进行的下一个最高唯一数字.
例如：
Symbols =[1,2,3,4,5] size = 3; given combination = 123,result = 124 given combination = 254,result = 312
解决方法
这是一个伪代码算法：
int n = length(Symbols);
int k = length(A);
// TRACK WHICH LETTERS ARE STILL AVAILABLE
available = sort(Symbols minus A);
// SEARCH BACKWARDS FOR AN ENTRY THAT CAN BE INCREASED
for (int i=k-1; i>=0; --i) {
// LOOK FOR NEXT SMALLEST AVAILABLE LETTER
for (int j=0; j<n-k; ++j) {
if (A[i] < available[j]) {
break;
}
}
if (j < n-k) {
// CHANGE A[i] TO THAT,REMOVE IT FROM AVAILABLE
int tmp = A[i];
A[i] = available[j];
available[j] = tmp;
// RESET SUBSEQUENT ENTRIES TO SMALLEST AVAILABLE
for (j=i+1; i<k; ++j) {
A[j] = available[i+1-j];
}
return A;
} else {
// A[i] MUST BE LARGER THAN AVAILABLE,SO APPEND TO END
available = append(available,A[i]);
}
}