序列子集(重复元素/无重复元素)

Author: cls1991

Backtracking/78_Subsets.py

@@ -0,0 +1,49 @@
+# coding: utf8
+
+
+"""
+    题目链接: https://leetcode.com/problems/subsets/description.
+    题目描述:
+
+    Given a set of distinct integers, nums, return all possible subsets (the power set).
+
+    Note: The solution set must not contain duplicate subsets.
+
+    For example,
+    If nums = [1,2,3], a solution is:
+
+    [
+      [3],
+      [1],
+      [2],
+      [1,2,3],
+      [1,3],
+      [2,3],
+      [1,2],
+      []
+    ]
+
+"""
+
+
+class Solution(object):
+    def subsets(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: List[List[int]]
+        """
+        if not nums:
+            return []
+
+        ans = []
+        sub = []
+        self.backtrace(nums, 0, sub, ans)
+
+        return ans
+
+    def backtrace(self, nums, idx, sub, ans):
+        ans.append(sub[:])
+        for i in range(idx, len(nums)):
+            sub.append(nums[i])
+            self.backtrace(nums, i + 1, sub, ans)
+            sub.pop()

Backtracking/90_SubsetsII.py

@@ -0,0 +1,54 @@
+# coding: utf8
+
+
+"""
+    题目链接: https://leetcode.com/problems/subsets-ii/description.
+    题目描述:
+
+    Given a collection of integers that might contain duplicates, nums, return all possible subsets (the power set).
+
+    Note: The solution set must not contain duplicate subsets.
+
+    For example,
+    If nums = [1,2,2], a solution is:
+
+    [
+      [2],
+      [1],
+      [1,2,2],
+      [2,2],
+      [1,2],
+      []
+    ]
+
+"""
+
+
+class Solution(object):
+    def subsetsWithDup(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: List[List[int]]
+        """
+        if not nums:
+            return []
+
+        ans = []
+        sub = []
+        visited = [0 for _ in range(len(nums))]
+        nums.sort()
+        self.backtrace(nums, 0, visited, sub, ans)
+
+        return ans
+
+    def backtrace(self, nums, idx, visited, sub, ans):
+        ans.append(sub[:])
+        for i in range(idx, len(nums)):
+            # 前面有重复元素未使用
+            if i > 0 and nums[i] == nums[i - 1] and not visited[i - 1]:
+                continue
+            visited[i] = 1
+            sub.append(nums[i])
+            self.backtrace(nums, i + 1, visited, sub, ans)
+            sub.pop()
+            visited[i] = 0

ProblemsList.md

@@ -196,6 +196,7 @@
 * 47\. [Permutations II](https://leetcode.com/problems/permutations-ii/description)
 * 51\. [N-Queens](https://leetcode.com/problems/n-queens/description)
 * 52\. [N-Queens II](https://leetcode.com/problems/n-queens-ii/description)
+* 78\. [Subsets](https://leetcode.com/problems/subsets/description)
 * 79\. [Word Search](https://leetcode.com/problems/word-search/description)
 * 90\. [Subsets II](https://leetcode.com/problems/subsets-ii/description)
 * 93\. [Restore IP Addresses](https://leetcode.com/problems/restore-ip-addresses/description)