数组全排列(重复元素/无重复元素)

Author: cls1991

Backtracking/46_Permutations.py

@@ -0,0 +1,50 @@
+# coding: utf8
+
+
+"""
+    题目链接: https://leetcode.com/problems/permutations/description.
+    题目描述:
+
+    Given a collection of distinct 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],
+          [3,2,1]
+        ]
+
+"""
+
+
+class Solution(object):
+    def permuteUnique(self, nums):
+        """
+        :type nums: List[int]
+        :rtype: List[List[int]]
+        """
+        if not nums:
+            return []
+
+        ans = []
+        sub = []
+        visited = [0 for _ in range(len(nums))]
+        self.backtrace(nums, visited, sub, ans)
+
+        return ans
+
+    def backtrace(self, nums, visited, sub, ans):
+        if len(sub) == len(nums):
+            ans.append(sub[:])
+        else:
+            for i in range(len(nums)):
+                if not visited[i]:
+                    visited[i] = 1
+                    sub.append(nums[i])
+                    self.backtrace(nums, visited, sub, ans)
+                    sub.pop()
+                    visited[i] = 0

Backtracking/47_PermutationsII.py

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+# coding: utf8
+
+
+"""
+    题目链接: https://leetcode.com/problems/permutations-ii/description.
+    题目描述:
+
+    Given a collection of numbers that might contain duplicates, return all possible unique permutations.
+
+    For example,
+        [1,1,2] have the following unique permutations:
+        [
+          [1,1,2],
+          [1,2,1],
+          [2,1,1]
+        ]
+
+"""
+
+
+class Solution(object):
+    def permuteUnique(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, visited, sub, ans)
+
+        return ans
+
+    def backtrace(self, nums, visited, sub, ans):
+        if len(sub) == len(nums):
+            ans.append(sub[:])
+        else:
+            for i in range(len(nums)):
+                if not visited[i]:
+                    # 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, visited, sub, ans)
+                    sub.pop()
+                    visited[i] = 0