组合排列: 全排列

Change-Id: I9df53cce6b2f82cabf728eec48744196bef2617f

Author: qieguo2016

Diff for: go/leetcode/46.全排列.go

@@ -0,0 +1,59 @@
+/*
+ * @lc app=leetcode.cn id=46 lang=golang
+ *
+ * [46] 全排列
+ *
+ * https://leetcode-cn.com/problems/permutations/description/
+ *
+ * algorithms
+ * Medium (71.53%)
+ * Likes:    365
+ * Dislikes: 0
+ * Total Accepted:    43.4K
+ * Total Submissions: 60.7K
+ * Testcase Example:  '[1,2,3]'
+ *
+ * 给定一个没有重复数字的序列，返回其所有可能的全排列。
+ * 
+ * 示例:
+ * 
+ * 输入: [1,2,3]
+ * 输出:
+ * [
+ * ⁠ [1,2,3],
+ * ⁠ [1,3,2],
+ * ⁠ [2,1,3],
+ * ⁠ [2,3,1],
+ * ⁠ [3,1,2],
+ * ⁠ [3,2,1]
+ * ]
+ * 
+ */
+
+
+func permute(nums []int) [][]int {
+	visited := make([]bool, len(nums))
+	out := []int{}
+	res := [][]int{}
+	permuteDFS(&nums, 0, &visited, &out, &res)
+	return res
+}
+
+// 深度优先递归，通过访问标记数组去重
+func permuteDFS(nums *[]int, level int, visited *[]bool, out *[]int, res *[][]int) {
+	if level == len(*nums) {
+		*res = append(*res, append([]int{}, (*out)...))  // 复制值
+		return 
+	}
+	for i := 0; i < len(*nums); i++ {
+		if (*visited)[i] {
+			continue
+		}
+		(*visited)[i] = true
+		*out = append(*out, (*nums)[i])
+		permuteDFS(nums, level+1, visited, out, res)
+		*out = (*out)[:len(*out)-1]
+		(*visited)[i] = false
+	}
+}
+