排列组合问题

一.不同元素子集问题

78. Subsets

Given a set of distinct integers, nums, return all possible subsets. 
给定一组非重复数字，求出所有可能的子集

解析：

例如 [1,2,3]，解法： 
首先放[]，然后往已有的[]中放1 
1. 首先放1 
此时已有[ [], 1 ] 
2. 然后对[ [], 1 ] 放2 
于是此时有 [ [], [1], [2], [1,2] ] 
3. 然后对[ [], [1], [2], [1,2] ] 放3 
此时[ [], [1], [2], [1,2], [3], [1,3], [2,3], [1,2,3] ] 求出解

#encoding:utf-8
_author_ = "Wang Wenchao"
def fun(ls):
    result=[]
    st=[]
    result.append(st)
    for i in ls:
        for j in range(len(result)):
            x=result[j][:] #复制列表
            x.append(i)
            result.append(x)
    return result
ls=[1,2,3]
print fun(ls)

　　运行结果    [[], [1], [2], [1, 2], [3], [1, 3], [2, 3], [1, 2, 3]]

二.全排列组合

46.Permutations(排列组合)

Permutations

Given a collection of numbers, return all possible permutations.

For example,
[1,2,3] have the following permutations:
[1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1].

 

 

#encoding:utf-8
_author_ = "Wang Wenchao"
count=0

def perm(ls,begin,end):
    global count
    if begin>=end:
        print ls
        count+=1
    else:
        i=begin
        for num in range(begin,end):
            ls[num],ls[i]=ls[i],ls[num]
            perm(ls,begin+1,end)
            ls[num], ls[i] = ls[i], ls[num]#回溯
ls = [1, 2,3]
perm(ls,0,len(ls))
print count

　结果：　

[1, 2, 3]
[1, 3, 2]
[2, 1, 3]
[2, 3, 1]
[3, 2, 1]
[3, 1, 2]
6

三.如果带有重复元素

则需要一个判断条件

#encoding:utf-8
_author_ = "Wang Wenchao"
count=0

def perm(ls,begin,end):
    global count
    if begin>=end:
        print ls
        count+=1
    else:
        i=begin
        for num in range(begin,end):
            if ls[num] not in ls[begin:num]:
                ls[num],ls[i]=ls[i],ls[num]
                perm(ls,begin+1,end)
                ls[num], ls[i] = ls[i], ls[num]#回溯
ls = [3,1,1,1]
perm(ls,0,len(ls))
print count

运行结果：　

[3, 1, 1, 1]
[1, 3, 1, 1]
[1, 1, 3, 1]
[1, 1, 1, 3]
4