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| 1 | +\chapter{暴力枚举法} |
| 2 | + |
| 3 | + |
| 4 | +\section{Subsets} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% |
| 5 | +\label{sec:subsets} |
| 6 | + |
| 7 | + |
| 8 | +\subsubsection{描述} |
| 9 | +Given a set of distinct integers, $S$, return all possible subsets. |
| 10 | + |
| 11 | +Note: |
| 12 | +\begindot |
| 13 | +\item Elements in a subset must be in non-descending order. |
| 14 | +\item The solution set must not contain duplicate subsets. |
| 15 | +\myenddot |
| 16 | + |
| 17 | +For example, If \code{S = [1,2,3]}, a solution is: |
| 18 | +\begin{Code} |
| 19 | +[ |
| 20 | + [3], |
| 21 | + [1], |
| 22 | + [2], |
| 23 | + [1,2,3], |
| 24 | + [1,3], |
| 25 | + [2,3], |
| 26 | + [1,2], |
| 27 | + [] |
| 28 | +] |
| 29 | +\end{Code} |
| 30 | + |
| 31 | + |
| 32 | +\subsection{增量构造法} |
| 33 | +每个元素,都有两种选择,选或者不选。 |
| 34 | + |
| 35 | +\subsubsection{代码} |
| 36 | +\begin{Code} |
| 37 | +// LeetCode, Subsets |
| 38 | +// 增量构造法,朴素深搜 |
| 39 | +class Solution { |
| 40 | +public: |
| 41 | + vector<vector<int> > subsets(vector<int> &S) { |
| 42 | + vector<vector<int> > result; |
| 43 | + vector<int> cur; |
| 44 | + std::sort(S.begin(), S.end()); // 本题对顺序有要求,需要排序 |
| 45 | + |
| 46 | + subsets(S, cur, 0, result); |
| 47 | + return result; |
| 48 | + } |
| 49 | + |
| 50 | +private: |
| 51 | + static void subsets(const vector<int> &S, vector<int> &cur, int step, |
| 52 | + vector<vector<int> > &result) { |
| 53 | + if (step == S.size()) { |
| 54 | + result.push_back(cur); |
| 55 | + return; |
| 56 | + } |
| 57 | + // 不选S[step] |
| 58 | + subsets(S, cur, step + 1, result); |
| 59 | + // 选S[step] |
| 60 | + cur.push_back(S[step]); |
| 61 | + subsets(S, cur, step + 1, result); |
| 62 | + cur.pop_back(); |
| 63 | + } |
| 64 | +}; |
| 65 | +\end{Code} |
| 66 | + |
| 67 | + |
| 68 | +\subsection{位向量法} |
| 69 | +开一个位向量\fn{bool selected[n]},每个元素可以选或者不选。 |
| 70 | + |
| 71 | + |
| 72 | +\subsubsection{代码} |
| 73 | +\begin{Code} |
| 74 | +// LeetCode, Subsets |
| 75 | +// 位向量法,也属于朴素深搜 |
| 76 | +class Solution { |
| 77 | +public: |
| 78 | + vector<vector<int> > subsets(vector<int> &S) { |
| 79 | + vector<vector<int> > result; |
| 80 | + bool *selected = new bool[S.size()]; |
| 81 | + std::sort(S.begin(), S.end()); // 本题对顺序有要求,需要排序 |
| 82 | + |
| 83 | + subsets(S, selected, 0, result); |
| 84 | + delete[] selected; |
| 85 | + return result; |
| 86 | + } |
| 87 | + |
| 88 | +private: |
| 89 | + static void subsets(const vector<int> &S, bool selected[], int step, |
| 90 | + vector<vector<int> > &result) { |
| 91 | + if (step == S.size()) { |
| 92 | + vector<int> subset; |
| 93 | + for (int i = 0; i < S.size(); i++) { |
| 94 | + if (selected[i]) subset.push_back(S[i]); |
| 95 | + } |
| 96 | + result.push_back(subset); |
| 97 | + return; |
| 98 | + } |
| 99 | + // 不选S[step] |
| 100 | + selected[step] = false; |
| 101 | + subsets(S, selected, step + 1, result); |
| 102 | + // 选S[step] |
| 103 | + selected[step] = true; |
| 104 | + subsets(S, selected, step + 1, result); |
| 105 | + } |
| 106 | +}; |
| 107 | +\end{Code} |
| 108 | + |
| 109 | + |
| 110 | +\subsection{二进制法} |
| 111 | +本方法的前提是:集合的元素不超过int位数。用一个int整数表示位向量,第$i$位为1,则表示选择$S[i]$,为0则不选择。例如\fn{S=\{A,B,C,D\}},则\fn{0110=6}表示子集\fn{\{B,C\}}。 |
| 112 | + |
| 113 | +这种方法最巧妙。因为它不仅能生成子集,还能方便的表示集合的并、交、差等集合运算。设两个集合的位向量分别为$B_1$和$B_2$,则$B_1|B_2, B_1 \& B_2, B_1 \^ B_2$分别对应集合的并、交、对称差。 |
| 114 | + |
| 115 | +\begin{Code} |
| 116 | +// LeetCode, Subsets |
| 117 | +// 二进制法 |
| 118 | +class Solution { |
| 119 | +public: |
| 120 | + vector<vector<int> > subsets(vector<int> &S) { |
| 121 | + vector<vector<int> > result; |
| 122 | + std::sort(S.begin(), S.end()); // 本题对顺序有要求,需要排序 |
| 123 | + const size_t n = S.size(); |
| 124 | + vector<int> v; |
| 125 | + |
| 126 | + for (size_t i = 0; i < 1 << n; i++) { |
| 127 | + for (size_t j = 0; j < n; j++) { |
| 128 | + if (i & 1 << j) v.push_back(S[j]); |
| 129 | + } |
| 130 | + result.push_back(v); |
| 131 | + v.clear(); |
| 132 | + } |
| 133 | + return result; |
| 134 | + } |
| 135 | +}; |
| 136 | +\end{Code} |
| 137 | + |
| 138 | + |
| 139 | +\subsubsection{代码} |
| 140 | +\begin{Code} |
| 141 | + |
| 142 | +\end{Code} |
| 143 | + |
| 144 | +\subsubsection{相关题目} |
| 145 | +\begindot |
| 146 | +\item Subsets II,见 \S \ref{sec:subsets-ii} |
| 147 | +\myenddot |
| 148 | + |
| 149 | + |
| 150 | +\section{Subsets II} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% |
| 151 | +\label{sec:subsets-ii} |
| 152 | + |
| 153 | + |
| 154 | +\subsubsection{描述} |
| 155 | +Given a collection of integers that might contain duplicates, $S$, return all possible subsets. |
| 156 | + |
| 157 | +Note: |
| 158 | + |
| 159 | +Elements in a subset must be in non-descending order. |
| 160 | +The solution set must not contain duplicate subsets. |
| 161 | +For example, |
| 162 | +If \fn{S = [1,2,2]}, a solution is: |
| 163 | +\begin{Code} |
| 164 | +[ |
| 165 | + [2], |
| 166 | + [1], |
| 167 | + [1,2,2], |
| 168 | + [2,2], |
| 169 | + [1,2], |
| 170 | + [] |
| 171 | +] |
| 172 | +\end{Code} |
| 173 | + |
| 174 | + |
| 175 | +\subsubsection{分析} |
| 176 | +这题有重复元素,但本质上,跟上一题很类似,上一题中元素没有重复,相当于每个元素只能选0或1次,这里扩充到了每个元素可以选0到若干次而已。 |
| 177 | + |
| 178 | + |
| 179 | +\subsubsection{代码} |
| 180 | +\begin{Code} |
| 181 | +// LeetCode, Subsets II |
| 182 | +class Solution { |
| 183 | +public: |
| 184 | + vector<vector<int> > subsetsWithDup(vector<int> &S) { |
| 185 | + vector<vector<int> > result; |
| 186 | + std::sort(S.begin(), S.end()); // 本题对顺序有要求,需要排序 |
| 187 | + std::vector<int> count(S.back() - S.front() + 1, 0); |
| 188 | + // 计算所有元素的个数 |
| 189 | + for(auto i = S.begin(); i != S.end(); i++) { |
| 190 | + count[*i - S[0]]++; |
| 191 | + } |
| 192 | + |
| 193 | + // 每个元素选择了多少个 |
| 194 | + std::vector<int> selected(S.back() - S.front() + 1, -1); |
| 195 | + |
| 196 | + subsets(S, count, selected, 0, result); |
| 197 | + return result; |
| 198 | + } |
| 199 | + |
| 200 | +private: |
| 201 | + static void subsets(const vector<int> &S, vector<int> &count, |
| 202 | + vector<int> &selected, size_t step, vector<vector<int> > &result) { |
| 203 | + if (step == count.size()) { |
| 204 | + vector<int> subset; |
| 205 | + for(size_t i = 0; i < selected.size(); i++) { |
| 206 | + for (int j = 0; j < selected[i]; j++) { |
| 207 | + subset.push_back(i+S[0]); |
| 208 | + } |
| 209 | + } |
| 210 | + result.push_back(subset); |
| 211 | + return; |
| 212 | + } |
| 213 | + |
| 214 | + for (int i = 0; i <= count[step]; i++) { |
| 215 | + selected[step] = i; |
| 216 | + subsets(S, count, selected, step + 1, result); |
| 217 | + } |
| 218 | + } |
| 219 | +}; |
| 220 | +\end{Code} |
| 221 | + |
| 222 | + |
| 223 | +\subsubsection{相关题目} |
| 224 | +\begindot |
| 225 | +\item Subsets,见 \S \ref{sec:subsets} |
| 226 | +\myenddot |
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