全排列

遞歸實現：把元素換到數組首位，剩下的部分做全排列

def constraint():return Truedef bound():return Truedef perm_backtracking(depth,lst):size = len(lst)if depth == size:print(lst)else:for i in range(depth,size): if constraint() and bound():lst[depth],lst[i] = lst[i],lst[depth]perm_backtracking(depth+1,lst)lst[depth],lst[i] = lst[i],lst[depth]

迭代實現，可以參考完全ｎ叉樹，這個應該是一個通用的迭代的處理方法，在這里：只要不在同一列就行了

def perm_backtracking_iteration(depth,lst):size = len(lst)Selected =[-1]*size# 排列數，列號不能重復def place(k):for i in range(k):if Selected[i] == Selected[k]:return Falsereturn Truewhile depth >=0:Selected[depth] +=1#　直到選擇一個可行的解while Selected[depth]<size and not place(depth):Selected[depth] +=1#　這個就是回溯條件，子集樹不為空if Selected[depth] <= size-1:if depth == size-1: # print Selected#　解碼結果，Selected里面對應的是選擇了哪一個數for i in Selected:print lst[i],print ""else:#　到達下一層時，初始化下一層的子樹的范圍depth +=1Selected[depth] =-1#　回溯，還是從上一層的未選的的地方走else:depth -=1A = ['A','B','C','D'] perm_backtracking(0,A)['A', 'B', 'C', 'D'] ['A', 'B', 'D', 'C'] ['A', 'C', 'B', 'D'] ['A', 'C', 'D', 'B'] ['A', 'D', 'C', 'B'] ['A', 'D', 'B', 'C'] ['B', 'A', 'C', 'D'] ['B', 'A', 'D', 'C'] ['B', 'C', 'A', 'D'] ['B', 'C', 'D', 'A'] ['B', 'D', 'C', 'A'] ['B', 'D', 'A', 'C'] ['C', 'B', 'A', 'D'] ['C', 'B', 'D', 'A'] ['C', 'A', 'B', 'D'] ['C', 'A', 'D', 'B'] ['C', 'D', 'A', 'B'] ['C', 'D', 'B', 'A'] ['D', 'B', 'C', 'A'] ['D', 'B', 'A', 'C'] ['D', 'C', 'B', 'A'] ['D', 'C', 'A', 'B'] ['D', 'A', 'C', 'B'] ['D', 'A', 'B', 'C']

還可以基于排列樹的迭代方式：

def perm_backtracking_iteration2(depth,lst):size = len(lst)Selected =[i for i in range(size)]change = [-1]*size count = 0while depth >=0:# range(Selected[depth],size)記錄的是剩余的次數，這里面還不能表現到底是那幾次？估計還可以改進if Selected[depth] < size:for i in range(Selected[depth],size):# 交換且記錄交換lst[depth],lst[i] = lst[i],lst[depth]change[depth] = iSelected[depth] +=1if depth == size-1:count +=1print lstprint countelse: depth +=1Selected[depth] = depthbreakelse:# 子集樹為空了，回溯到上一層,在最底部是沒有交換動作的，因為change[depth]就是本身，# 所以需要先-1，再交換，可以自己先調試一遍depth -=1 lst[depth],lst[change[depth]] = lst[change[depth]],lst[depth]change[depth] = -1A = ['A','B','C','D'] #perm_backtracking(0,A) perm_backtracking_iteration2(0,A)['A', 'B', 'C', 'D'] 1 ['A', 'B', 'D', 'C'] 2 ['A', 'C', 'B', 'D'] 3 ['A', 'C', 'D', 'B'] 4 ['A', 'D', 'C', 'B'] 5 ['A', 'D', 'B', 'C'] 6 ['B', 'A', 'C', 'D'] 7 ['B', 'A', 'D', 'C'] 8 ['B', 'C', 'A', 'D'] 9 ['B', 'C', 'D', 'A'] 10 ['B', 'D', 'C', 'A'] 11 ['B', 'D', 'A', 'C'] 12 ['C', 'B', 'A', 'D'] 13 ['C', 'B', 'D', 'A'] 14 ['C', 'A', 'B', 'D'] 15 ['C', 'A', 'D', 'B'] 16 ['C', 'D', 'A', 'B'] 17 ['C', 'D', 'B', 'A'] 18 ['D', 'B', 'C', 'A'] 19 ['D', 'B', 'A', 'C'] 20 ['D', 'C', 'B', 'A'] 21 ['D', 'C', 'A', 'B'] 22 ['D', 'A', 'C', 'B'] 23 ['D', 'A', 'B', 'C'] 24

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