80 lines
1.9 KiB
Java
80 lines
1.9 KiB
Java
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//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。
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//
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//
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//
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// 示例 1：
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//
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//
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//输入：nums = [1,1,2]
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//输出：
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//[[1,1,2],
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// [1,2,1],
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// [2,1,1]]
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//
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//
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// 示例 2：
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//
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//
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//输入：nums = [1,2,3]
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//输出：[[1,2,3],[1,3,2],[2,1,3],[2,3,1],[3,1,2],[3,2,1]]
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//
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//
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//
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//
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// 提示：
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//
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//
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// 1 <= nums.length <= 8
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// -10 <= nums[i] <= 10
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//
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// Related Topics 回溯算法
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// 👍 681 👎 0
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package leetcode.editor.cn;
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import java.util.ArrayList;
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import java.util.Arrays;
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import java.util.List;
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//47:全排列 II
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public class PermutationsIi {
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public static void main(String[] args) {
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//测试代码
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Solution solution = new PermutationsIi().new Solution();
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solution.permuteUnique(new int[]{1, 1, 2});
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}
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//力扣代码
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//leetcode submit region begin(Prohibit modification and deletion)
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class Solution {
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boolean[] vis;
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public List<List<Integer>> permuteUnique(int[] nums) {
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List<List<Integer>> ans = new ArrayList<List<Integer>>();
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List<Integer> perm = new ArrayList<Integer>();
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vis = new boolean[nums.length];
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Arrays.sort(nums);
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backtrack(nums, ans, 0, perm);
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return ans;
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}
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public void backtrack(int[] nums, List<List<Integer>> ans, int idx, List<Integer> perm) {
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if (idx == nums.length) {
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ans.add(new ArrayList<Integer>(perm));
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return;
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}
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for (int i = 0; i < nums.length; ++i) {
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if (vis[i] || (i > 0 && nums[i] == nums[i - 1] && !vis[i - 1])) {
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continue;
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}
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perm.add(nums[i]);
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vis[i] = true;
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backtrack(nums, ans, idx + 1, perm);
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vis[i] = false;
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perm.remove(idx);
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}
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}
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}
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//leetcode submit region end(Prohibit modification and deletion)
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}
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