C++实现LeetCode(90.子集合之二)

[LeetCode] 90. Subsets II 子集合之二

Given a collection of integers that might contain duplicates, S, return all possible subsets.

Note:

  • Elements in a subset must be in non-descending order.
  • The solution set must not contain duplicate subsets.

For example,
If S = [1,2,2], a solution is:

[
[2],
[1],
[1,2,2],
[2,2],
[1,2],
[]
]

这道子集合之二是之前那道 Subsets 的延伸，这次输入数组允许有重复项，其他条件都不变，只需要在之前那道题解法的基础上稍加改动便可以做出来，我们先来看非递归解法，拿题目中的例子 [1 2 2] 来分析，根据之前 Subsets 里的分析可知，当处理到第一个2时，此时的子集合为 [], [1], [2], [1, 2]，而这时再处理第二个2时，如果在 [] 和 [1] 后直接加2会产生重复，所以只能在上一个循环生成的后两个子集合后面加2，发现了这一点，题目就可以做了，我们用 last 来记录上一个处理的数字，然后判定当前的数字和上面的是否相同，若不同，则循环还是从0到当前子集的个数，若相同，则新子集个数减去之前循环时子集的个数当做起点来循环，这样就不会产生重复了，代码如下：

解法一：

class Solution {
public:
    vector<vector<int>> subsetsWithDup(vector<int> &S) {
        if (S.empty()) return {};
        vector<vector<int>> res(1);
        sort(S.begin(), S.end());
        int size = 1, last = S[0];
        for (int i = 0; i < S.size(); ++i) {
            if (last != S[i]) {
                last = S[i];
                size = res.size();
            }
            int newSize = res.size();
            for (int j = newSize - size; j < newSize; ++j) {
                res.push_back(res[j]);
                res.back().push_back(S[i]);
            }
        }
        return res;
    }
};

整个添加的顺序为：

[]
[1]
[2]
[1 2]
[2 2]
[1 2 2]

对于递归的解法，根据之前 Subsets 里的构建树的方法，在处理到第二个2时，由于前面已经处理了一次2，这次我们只在添加过2的 [2] 和 [1 2] 后面添加2，其他的都不添加，那么这样构成的二叉树如下图所示：

                        []
                   /          \
                  /            \
                 /              \
              [1]                []
           /       \           /    \
          /         \         /      \
       [1 2]       [1]       [2]     []
      /     \     /   \     /   \    / \
  [1 2 2] [1 2]  X   [1]  [2 2] [2] X  []

代码只需在原有的基础上增加一句话，while (S[i] == S[i + 1]) ++i; 这句话的作用是跳过树中为X的叶节点，因为它们是重复的子集，应被抛弃。代码如下：

解法二：

class Solution {
public:
    vector<vector<int>> subsetsWithDup(vector<int> &S) {
        if (S.empty()) return {};
        vector<vector<int>> res;
        vector<int> out;
        sort(S.begin(), S.end());
        getSubsets(S, 0, out, res);
        return res;
    }
    void getSubsets(vector<int> &S, int pos, vector<int> &out, vector<vector<int>> &res) {
        res.push_back(out);
        for (int i = pos; i < S.size(); ++i) {
            out.push_back(S[i]);
            getSubsets(S, i + 1, out, res);
            out.pop_back();
            while (i + 1 < S.size() && S[i] == S[i + 1]) ++i;
        }
    }
};

整个添加的顺序为：

[]
[1]
[1 2]
[1 2 2]
[2]
[2 2]

到此这篇关于C++实现LeetCode(90.子集合之二)的文章就介绍到这了,更多相关C++实现子集合之二内容请搜索我们以前的文章或继续浏览下面的相关文章希望大家以后多多支持我们！

赞 (0)

相关推荐

  • C++实现LeetCode(78.子集合)

    [LeetCode] 78. Subsets 子集合 Given a set of distinct integers, S, return all possible subsets. Note: Elements in a subset must be in non-descending order. The solution set must not contain duplicate subsets. For example, If S = [1,2,3], a solution is:

  • C++实现LeetCode(51.N皇后问题)

    C++实现LeetCode(51.N皇后问题)

    [LeetCode] 51. N-Queens N皇后问题 The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other. Given an integer n, return all distinct solutions to the n-queens puzzle. Each solution contains a di

  • C++实现LeetCode(46.全排列)

    [LeetCode] 46. Permutations 全排列 Given a collection of distinct integers, return all possible permutations. Example: Input: [1,2,3] Output: [ [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1] ] 这道题是求全排列问题,给的输入数组没有重复项,这跟之前的那道 Combinations 和类似,解法基本相同

  • C++实现LeetCode(144.二叉树的先序遍历)

    [LeetCode] 144. Binary Tree Preorder Traversal 二叉树的先序遍历 Given a binary tree, return the preorder traversal of its nodes' values. Example: Input:  [1,null,2,3] 1 \ 2 / 3 Output:  [1,2,3] Follow up: Recursive solution is trivial, could you do it iterat

  • C++实现LeetCode(112.二叉树的路径和)

    [LeetCode] 112. Path Sum 二叉树的路径和 Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum. Note: A leaf is a node with no children. Example: Given the below bi

  • C++实现LeetCode(38.计数和读法)

    [LeetCode] 38. Count and Say 计数和读法 The count-and-say sequence is the sequence of integers with the first five terms as following: 1.     1 2.     11 3.     21 4.     1211 5.     111221 1 is read off as "one 1" or 11. 11 is read off as "two

  • C++实现LeetCode(39.组合之和)

    [LeetCode] 39. Combination Sum 组合之和 Given a set of candidate numbers (candidates) (without duplicates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target. The same repeated number may b

  • C++实现LeetCode(113.二叉树路径之和之二)

    [LeetCode] 113. Path Sum II 二叉树路径之和之二 Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. For example: Given the below binary tree and sum = 22,  5 / \ 4   8 /      / \ 11  13  4 /  \         / \ 7  

  • C++实现LeetCode(90.子集合之二)

    [LeetCode] 90. Subsets II 子集合之二 Given a collection of integers that might contain duplicates, S, return all possible subsets. Note: Elements in a subset must be in non-descending order. The solution set must not contain duplicate subsets. For example

  • C++实现LeetCode(45.跳跃游戏之二)

    [LeetCode] 45. Jump Game II 跳跃游戏之二 Given an array of non-negative integers, you are initially positioned at the first index of the array. Each element in the array represents your maximum jump length at that position. Your goal is to reach the last i

  • C++实现LeetCode(40.组合之和之二)

    [LeetCode] 40. Combination Sum II 组合之和之二 Given a collection of candidate numbers (candidates) and a target number (target), find all unique combinations in candidates where the candidate numbers sums to target. Each number in candidates may only be u

  • C++实现LeetCode(59.螺旋矩阵之二)

    [LeetCode] 59. Spiral Matrix II 螺旋矩阵之二 Given a positive integer n, generate a square matrix filled with elements from 1 to n2 in spiral order. Example: Input: 3 Output: [ [ 1, 2, 3 ], [ 8, 9, 4 ], [ 7, 6, 5 ] ] 此题跟之前那道 Spiral Matrix 本质上没什么区别,就相当于个类似逆

  • C++实现LeetCode(126.词语阶梯之二)

    [LeetCode] 126. Word Ladder II 词语阶梯之二 Given two words (beginWord and endWord), and a dictionary's word list, find all shortest transformation sequence(s) from beginWord to endWord, such that: Only one letter can be changed at a time Each transformed

  • C++实现LeetCode(140.拆分词句之二)

    [LeetCode] 140.Word Break II 拆分词句之二 Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences. Not

  • C++实现LeetCode(210.课程清单之二)

    [LeetCode] 210. Course Schedule II 课程清单之二 There are a total of n courses you have to take, labeled from 0 to n-1. Some courses may have prerequisites, for example to take course 0 you have to first take course 1, which is expressed as a pair: [0,1] G

  • C#归并排序的实现方法(递归,非递归,自然归并)

    C#归并排序的实现方法(递归,非递归,自然归并)

    //Main: 复制代码 代码如下: using System;using System.Collections.Generic;using System.Linq;using System.Text; namespace Merge{    class Program    {        static void Main(string[] args)        {            while (true)            {                Console.W

  • MongoDB教程之入门基础知识

    一.文档的注意事项: 1.  键值对是有序的,如:{ "name" : "stephen", "genda" : "male" } 不等于 { "genda" : "male", "name" : "stephen" } 2.  文档信息是大小写敏感的,如:{ "name" : "stephen" } 不等于

随机推荐