LeetCode | Subsets
2014-03-05 00:04
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题目
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 =
is:
分析
有n个元素的集合的真子集有2^n个,在此基础上增加1个元素,真子集个数变成了2^(n+1),这多出来的2^n个真子集就是在原来2^n个真子集上都添加一个新元素得到的。按照这个思路得到解法1。
如果元素个数小于32个,还可以采用解法2的思路:把n个元素对应到n个比特位上,那么穷举真子集其实就是遍历0到2^n-1这2^n个数字的二进制表示,比特位为1表示该位对应的元素存在。
解法1
import java.util.ArrayList;
import java.util.Arrays;
public class Subsets {
public ArrayList<ArrayList<Integer>> subsets(int[] S) {
Arrays.sort(S);
ArrayList<ArrayList<Integer>> results = new ArrayList<ArrayList<Integer>>();
ArrayList<Integer> list = new ArrayList<Integer>();
results.add(list);
for (int i = 0; i < S.length; ++i) {
int j = results.size();
while (j-- > 0) {
list = new ArrayList<Integer>(results.get(j));
list.add(S[i]);
results.add(list);
}
}
return results;
}
}解法2
import java.util.ArrayList;
import java.util.Arrays;
public class Subsets {
public ArrayList<ArrayList<Integer>> subsets(int[] S) {
ArrayList<ArrayList<Integer>> results = new ArrayList<ArrayList<Integer>>();
Arrays.sort(S);
int N = S.length;
for (int i = 0; i < Math.pow(2, N); ++i) {
ArrayList<Integer> list = new ArrayList<Integer>();
for (int j = 0; j < N; ++j) {
if ((i & (1 << j)) > 0) {
list.add(S[j]);
}
}
results.add(list);
}
return results;
}
}
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:
[ [3], [1], [2], [1,2,3], [1,3], [2,3], [1,2], [] ]
分析
有n个元素的集合的真子集有2^n个,在此基础上增加1个元素,真子集个数变成了2^(n+1),这多出来的2^n个真子集就是在原来2^n个真子集上都添加一个新元素得到的。按照这个思路得到解法1。
如果元素个数小于32个,还可以采用解法2的思路:把n个元素对应到n个比特位上,那么穷举真子集其实就是遍历0到2^n-1这2^n个数字的二进制表示,比特位为1表示该位对应的元素存在。
解法1
import java.util.ArrayList;
import java.util.Arrays;
public class Subsets {
public ArrayList<ArrayList<Integer>> subsets(int[] S) {
Arrays.sort(S);
ArrayList<ArrayList<Integer>> results = new ArrayList<ArrayList<Integer>>();
ArrayList<Integer> list = new ArrayList<Integer>();
results.add(list);
for (int i = 0; i < S.length; ++i) {
int j = results.size();
while (j-- > 0) {
list = new ArrayList<Integer>(results.get(j));
list.add(S[i]);
results.add(list);
}
}
return results;
}
}解法2
import java.util.ArrayList;
import java.util.Arrays;
public class Subsets {
public ArrayList<ArrayList<Integer>> subsets(int[] S) {
ArrayList<ArrayList<Integer>> results = new ArrayList<ArrayList<Integer>>();
Arrays.sort(S);
int N = S.length;
for (int i = 0; i < Math.pow(2, N); ++i) {
ArrayList<Integer> list = new ArrayList<Integer>();
for (int j = 0; j < N; ++j) {
if ((i & (1 << j)) > 0) {
list.add(S[j]);
}
}
results.add(list);
}
return results;
}
}
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