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#yyds干货盘点# leetcode算法题：全排列 II
source link: https://blog.51cto.com/u_13321676/5449314
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#yyds干货盘点# leetcode算法题：全排列 II
原创给定一个可包含重复数字的序列 nums ，按任意顺序 返回所有不重复的全排列。
输入：nums = [1,1,2]
[[1,1,2],
[1,2,1],
[2,1,1]]
输入：nums = [1,2,3]
输出：[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
代码实现：
class Solution {
boolean[] vis;
public List<List<Integer>> permuteUnique(int[] nums) {
List<List<Integer>> ans = new ArrayList<List<Integer>>();
List<Integer> perm = new ArrayList<Integer>();
vis = new boolean[nums.length];
Arrays.sort(nums);
backtrack(nums, ans, 0, perm);
return ans;
}
public void backtrack(int[] nums, List<List<Integer>> ans, int idx, List<Integer> perm) {
if (idx == nums.length) {
ans.add(new ArrayList<Integer>(perm));
return;
}
for (int i = 0; i < nums.length; ++i) {
if (vis[i] || (i > 0 && nums[i] == nums[i - 1] && !vis[i - 1])) {
continue;
}
perm.add(nums[i]);
vis[i] = true;
backtrack(nums, ans, idx + 1, perm);
vis[i] = false;
perm.remove(idx);
}
}
}
boolean[] vis;
public List<List<Integer>> permuteUnique(int[] nums) {
List<List<Integer>> ans = new ArrayList<List<Integer>>();
List<Integer> perm = new ArrayList<Integer>();
vis = new boolean[nums.length];
Arrays.sort(nums);
backtrack(nums, ans, 0, perm);
return ans;
}
public void backtrack(int[] nums, List<List<Integer>> ans, int idx, List<Integer> perm) {
if (idx == nums.length) {
ans.add(new ArrayList<Integer>(perm));
return;
}
for (int i = 0; i < nums.length; ++i) {
if (vis[i] || (i > 0 && nums[i] == nums[i - 1] && !vis[i - 1])) {
continue;
}
perm.add(nums[i]);
vis[i] = true;
backtrack(nums, ans, idx + 1, perm);
vis[i] = false;
perm.remove(idx);
}
}
}
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