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博客园 - Zero Lee

调用栈(call stack) 关于STL allocator Calculate maximum sum of any subarray set Calcuate power n of x recursively Convert one binary search tree to double-linked list 设计包含min函数的栈 类模板的模板友元函数定义 一道百度的面试题解答 非printf形式的十六进制和二进制打印(雅虎面试题) 一道腾讯面试题 (转)C++中extern “C”含义深层探索 selection algorithm to select nth small elements based on partition 删除与某个字符相邻且相同的字符 一组数的全排列和组合程序实现 求一个正整数的平方根程序实现 [转]多线程队列的算法优化 [转载] STL allocator的介绍和一个基于malloc/free的allocator的简单实现 如何将一片内存链接成链表 One simple counted object pointer
产生全排列的方法解析
Zero Lee · 2012-06-17 · via 博客园 - Zero Lee

A permutation can be obtained by selecting an element in the given set and recursively permuting the remaining elements.

 { ai,P(a1,...,ai-1,ai+1,...,aN) if N > 1 P(a1,...,aN) = aN if N = 1

 --|--|--|-| |a|b-|c-d-| a|------------b------------c-------------d --|--|--|-| ---|-|--|--| ---|--|-|--| --|--|--|-| |-|b-|c-d-| |a-|-|c-|d-| |a-|b-|-|d-| |a|b-|c-|-|

At each stage of the permutation process, the given set of elements consists of two parts: a subset of values that already have been processed, and a subset that still needs to be processed. This logical seperation can be physically realized by exchanging, in the i’th step, the i’th value with the value being chosen at that stage. That approaches leaves the first subset in the first i locations of the outcome.

 --|--|--|-| |a|b-|c-d-| --|--|------------|--------------------------|--|-| a||b |c d | |b a |c |d | |c||b |a|d | |d|b |c |a| -----|--------------------------|---- ----------- --|--|--|-| ---|-|--|--| ---|--|-|--| b-|a-|c-d-| |b-|c|a-|d-| |b-|d-|c|a-| ---|--|------------|--|-| |b-|c-a-|d-| b-|c-|d-|a| | b-|c-|d-|a| |-|--|--|-|

1 permute(i) 
2    if i == N  output A[N] 
3    else 
4       for j = i to N do 
5          swap(A[i], A[j]) 
6          permute(i+1) 
7          swap(A[i], A[j])