LeetCode#46 Permutations

Problem Definition:

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].

Solution:

1）借用 Next Permutation 算法（线性），从最小的开始，找下一个......

    # @param {integer[]} nums
    # @return {integer[][]}
    def permute(self, nums):
        nums.sort()
        pm=[nums[:]]
        cnt=reduce(lambda x, y: x*y, range(1, len(nums)+1))
        for _ in range(cnt-1):
            self.nextPermute(nums),
            pm+=nums[:],
        return pm

    def nextPermute(self, nums):
        n=len(nums)
        if n<2:
            return
        i,j,pLeft,pRight=n-2,n-1,-1,n-1
        while i>=0 and nums[i]>=nums[i+1]:
            i-=1
        if i>=0:
            while j>i and nums[j]<=nums[i]:
                j-=1
            #must be j>i
            nums[i],nums[j]=nums[j],nums[i]
        pLeft=i+1
        while pLeft<pRight:
            nums[pLeft],nums[pRight]=nums[pRight],nums[pLeft]
            pLeft+=1
            pRight-=1

2）上面的解法，实现nextPerte函数有很多的细节要考虑。这题用回溯法比较好，idea比较直接，实现起来容易，性能也好。

    # @param {integer[]} nums
    # @return {integer[][]}
    def permute(nums):
        pms=[]
        backTrack(nums, [], pms, len(nums))
        return pms

    def backTrack(nums, localArr, pms, n):
        if len(localArr)==n:
            pms+=localArr[:],
        else:
            for i, e in enumerate(nums):
                newNums=nums[:i]+nums[i+1:]
                backTrack(nums[:i]+nums[i+1:], localArr+[e], pms, n)