[Vssue]0090.子集II.md

[youngyangyang04]

https://www.programmercarl.com/0090.%E5%AD%90%E9%9B%86II.html

[Du1in9]

private void backtracking(int[] nums, int start) {
    paths.add(new ArrayList<>(path));
    for (int i = start; i < nums.length; i++) {
        if (i > start && nums[i] == nums[i - 1]) {
            continue;
        }
        path.add(nums[i]);
        backtracking(nums, i + 1);
        path.remove(path.size() - 1);
    }
}

// 深度1: backtracking([1,2,2], 0), path = []
i = 0：path = [1]，递归 backtracking([1,2,3], 1)，加入 paths。回溯，path = []
i = 1：path = [2]，递归 backtracking([1,2,3], 2)，加入 paths。回溯，path = []
i = 2：满足 nums[i] == nums[i - 1], 继续遍历（去重）
// 深度2: 例: backtracking([1,2,2], 1), path = [1]
i = 1：path = [1,2]，递归 backtracking([1,2,3], 2)，加入 paths。回溯，path = [1]
i = 2：满足 nums[i] == nums[i - 1], 继续遍历（去重）
// 深度3: 例: backtracking([1,2,2], 2), path = [1,2]
i = 2：path = [1,2,2]，递归 backtracking([1,2,3], 3)，加入 paths。回溯，path = [1,2]

[ZXiaoheng]

没必要用used进行记录。

class Solution:
    def subsetsWithDup(self, nums):
        result = []
        path = []
        nums.sort()
        self.backtracking(nums, 0,  path, result)
        return result

    def backtracking(self, nums, startIndex, path, result):
        result.append(path[:])  # 收集子集
        for i in range(startIndex, len(nums)):
            # used[i - 1] == True，说明同一树枝 nums[i - 1] 使用过
            # used[i - 1] == False，说明同一树层 nums[i - 1] 使用过
            # 而我们要对同一树层使用过的元素进行跳过
            if i > startIndex and nums[i] == nums[i - 1]:
                continue
            path.append(nums[i])
            self.backtracking(nums, i + 1, path, result)
            path.pop()