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]);
- }
- }
