标签：des blog io ar os sp java for div
Given a set of distinct integers, S, return all possible subsets.
Note:
For example,
If S = [1,2,3], a solution is:
[ [3], [1], [2], [1,2,3], [1,3], [2,3], [1,2], [] ]
1,解法：对于数组中的每一个值，都添加到原来的list中去。
2， arrsys本身的sort函数，满足题目中non-descending的要求
public class Solution {
public ArrayList<ArrayList<Integer>> subsets(int[] S) {
ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
if (S == null || S.length == 0) {
return result;
}
Arrays.sort(S);
result.add(new ArrayList<Integer>());
for (int i = 0; i < S.length; i++) {
int lengt = result.size();
for (int j = 0; j < lengt; j++) {
ArrayList<Integer> ad = new ArrayList<Integer>(result.get(j));
ad.add(S[i]);
result.add(ad);
}
}
return result;
}
}
标签：des blog io ar os sp java for div
原文地址：http://www.cnblogs.com/lilyfindjobs/p/4104823.html
