46. Permutations

#46. Permutations
def backtrackPerM(nums, visited, temp, res):
    # teminal condition
    if len(temp) == len(nums):
        res.append(temp)
        return
    # backtracking
    for i in range(0, len(nums)):
        if visited[i] == 0:
            visited[i] = 1      #每次到下层遍历时候要把当前所访问的数值 visited[i] = 1
            backtrackPerM(nums, visited, temp+[nums[i]], res)
            visited[i] = 0      #返回到当前曾以后再将 visited[i] = 0
    return

def permute(nums):
    res = []
    visited = [0] * len(nums) # 这道题要从头便利所有在数组中的数值。所以，要维护一个visited数组来记录当前值
    backtrackPerM(nums, visited, [], res)
    return res