題目

子集?

給定一個含不同整數的集合，返回其所有的子集

樣例

如果 S =?[1,2,3]，有如下的解：

[[3],[1],[2],[1,2,3],[1,3],[2,3],[1,2],[] ] 注意

子集中的元素排列必須是非降序的，解集必須不包含重復的子集

挑戰

你可以同時用遞歸與非遞歸的方式解決么？

解題

根據上面求排列的思想很類似，還是深度優先遍歷。由于輸出每個子集需要升序，所以要先對數組進行排序。求出所以的子集，也就是求出所以的組合方式 + 空集

問題轉化為求組合方式的問題

參考鏈接不僅要考慮起始位置，還需要考慮長度，這樣才是組合 C(n,k)，由于我只想到要考慮起始位置，而長度問題在程序中增加，一直沒有解決問題

核心程序

public void helper(int[] nums,int start,int len,ArrayList<Integer> list,ArrayList<ArrayList<Integer>> res){if( list.size() == len){res.add(new ArrayList<Integer>(list));return;}for(int i=start;i< nums.length;i++){if(list.contains(nums[i])){continue;}list.add(nums[i]);helper(nums,i+1,len,list,res);list.remove(list.size()-1);}} class Solution {/*** @param S: A set of numbers.* @return: A list of lists. All valid subsets.*/public ArrayList<ArrayList<Integer>> subsets(int[] nums) {// write your code hereArrayList<Integer> list = new ArrayList<Integer>();ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();if(nums == null || nums.length ==0)return res;Arrays.sort(nums);res.add(list);for(int len = 1;len<= nums.length;len++){helper(nums,0,len,list,res);}return res;}public void helper(int[] nums,int start,int len,ArrayList<Integer> list,ArrayList<ArrayList<Integer>> res){if( list.size() == len){res.add(new ArrayList<Integer>(list));return;}for(int i=start;i< nums.length;i++){if(list.contains(nums[i])){continue;}list.add(nums[i]);helper(nums,i+1,len,list,res);list.remove(list.size()-1);}} } Java Code

九章中程序進行了優化，長度不考慮，遞歸一次list的值都是子集的一個元素

class Solution {/*** @param S: A set of numbers.* @return: A list of lists. All valid subsets.*/public ArrayList<ArrayList<Integer>> subsets(int[] nums) {// write your code hereArrayList<Integer> list = new ArrayList<Integer>();ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();if(nums == null || nums.length ==0)return res;Arrays.sort(nums);helper(nums,0,list,res);return res;}public void helper(int[] nums,int start, ArrayList<Integer> list,ArrayList<ArrayList<Integer>> res){res.add(new ArrayList<Integer>(list));for(int i=start;i< nums.length;i++){if(list.contains(nums[i])){continue;}list.add(nums[i]);helper(nums,i+1,list,res);list.remove(list.size()-1);}} } Java Code class Solution:"""@param S: The set of numbers.@return: A list of lists. See example."""def subsets(self, S):def dfs(depth, start, valuelist):res.append(valuelist)if depth == len(S): returnfor i in range(start, len(S)):dfs(depth+1, i+1, valuelist+[S[i]])S.sort()res = []dfs(0, 0, [])return res Python Code

根據位運算進行求解

// 1 << n is 2^n// each subset equals to an binary integer between 0 .. 2^n - 1// 0 -> 000 -> []// 1 -> 001 -> [1]// 2 -> 010 -> [2]// ..// 7 -> 111 -> [1,2,3]

?下面是我自己實現

class Solution {/*** @param S: A set of numbers.* @return: A list of lists. All valid subsets.*/public ArrayList<ArrayList<Integer>> subsets(int[] nums) {// write your code hereint len = nums.length;ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();if(nums == null || len==0)return res;Arrays.sort(nums);// 1 << n is 2^n// each subset equals to an binary integer between 0 .. 2^n - 1// 0 -> 000 -> []// 1 -> 001 -> [1]// 2 -> 010 -> [2]// ..// 7 -> 111 -> [1,2,3]for(int i=0;i< 1<<len ;i++){ArrayList<Integer> list = new ArrayList<Integer>();// 檢測哪一位是 1 int n = i;for(int j=0;j< len;j++){if(n%2==1)list.add(nums[j]);n=n/2;}res.add(list);}return res;} } View Code

九章中判斷第幾位是1的程序如下：

for (int i = 0; i < (1 << n); i++) {ArrayList<Integer> subset = new ArrayList<Integer>();for (int j = 0; j < n; j++) {// check whether the jth digit in i's binary representation is 1if ((i & (1 << j)) != 0) {subset.add(nums[j]);}}result.add(subset);}

是懂非懂，好像這樣也可以判斷第幾位是1

class Solution:"""@param S: The set of numbers.@return: A list of lists. See example."""def subsets(self, nums):# write your code hereres = []size = len(nums)nums.sort()if nums == None or size == 0:return resfor i in range(1<<size):lst=[]n = ifor j in range(size):if n%2==1:lst.append(nums[j])n/=2res.append(lst)return res Python Code

?

?

轉載于:https://www.cnblogs.com/theskulls/p/5099745.html

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