生成不需要替换的置换数组

def permute(nums):
    def backtrack(first=0):
        if first == n:
            output.append(nums[:])
        for i in range(first, n):
            nums[first], nums[i] = nums[i], nums[first]
            backtrack(first + 1)
            nums[first], nums[i] = nums[i], nums[first]

    n = len(nums)
    output = []
    backtrack()
    return output

# 示例
nums = [1, 2, 3]
print(permute(nums))

def permuteUnique(nums):
    def backtrack(start=0):
        if start == len(nums):
            output.append(nums[:])
        used = set()
        for i in range(start, len(nums)):
            if nums[i] in used:
                continue
            used.add(nums[i])
            nums[start], nums[i] = nums[i], nums[start]
            backtrack(start + 1)
            nums[start], nums[i] = nums[i], nums[start]

    output = []
    nums.sort()
    backtrack(0)
    return output

# 示例
nums = [1, 1, 2]
print(permuteUnique(nums))