11. 递增子序列

var findSubsequences = function(nums) {let result = []let path = []function backtracing(startIndex) {if(path.length > 1) {result.push(path.slice())}let uset = []for(let i = startIndex; i < nums.length; i++) {if((path.length > 0 && nums[i] < path[path.length - 1]) || uset[nums[i] + 100]) {continue}uset[nums[i] + 100] = truepath.push(nums[i])backtracing(i + 1)path.pop()}}backtracing(0)return result
};

12. 全排列

const permute = (nums) => {const res = [];const used = {};function dfs(path) {if (path.length == nums.length) { // 个数选够了res.push(path.slice()); // 拷贝一份path，加入解集resreturn;                 // 结束当前递归分支}for (const num of nums) { // for枚举出每个可选的选项// if (path.includes(num)) continue; // 别这么写！查找是O(n)，增加时间复杂度if (used[num]) continue; // 使用过的，跳过path.push(num);         // 选择当前的数，加入pathused[num] = true;       // 记录一下 使用了dfs(path);              // 基于选了当前的数，递归path.pop();             // 上一句的递归结束，回溯，将最后选的数pop出来used[num] = false;      // 撤销这个记录}}dfs([]); // 递归的入口，空path传进去return res;
};

13. 全排列 II

var permuteUnique = function(nums) {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;
};