全排列 II
- https://leetcode.cn/problems/permutations-ii/description/
描述
- 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
示例 1
输入：nums = [1,1,2]
输出：
[[1,1,2],[1,2,1],[2,1,1]]
示例 2
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]]
提示
- 1 <= nums.length <= 8
- -10 <= nums[i] <= 10
Typescript 版算法实现
1 ) 方案1：搜索回溯
function permuteUnique(nums: number[]): number[][] {// 数字有重复，排列不能重复nums.sort((a, b) => b - a)let ret = []let path = []const backtrack = (used) => {if (path.length === nums.length) {ret.push([...path])return}for (let i = 0; i < nums.length; i++) {let num = nums[i]if (nums[i] === nums[i - 1] && !used[i - 1]) {continue}if (!used[i]) {used[i] = truepath.push(num)backtrack(used)path.pop()used[i] = false}}}backtrack([])return ret
};
2 ) 方案2：搜索回溯-优化版
function permuteUnique(nums: number[]): number[][] {const ans = [];const vis = new Array(nums.length).fill(false);const backtrack = (idx, perm) => {if (idx === nums.length) {ans.push(perm.slice());return;}for (let i = 0; i < nums.length; ++i) {if (vis[i] || (i > 0 && nums[i] === nums[i - 1] && !vis[i - 1])) {continue;}perm.push(nums[i]);vis[i] = true;backtrack(idx + 1, perm);vis[i] = false;perm.pop();}}nums.sort((x, y) => x - y);backtrack(0, []);return ans;
};
