【LeetCode】Subsets II 解题报告

【题目】

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],
  []
]

【解析】

看到这个题目，我立马想起了 【LeetCode】Combinations 解题报告 和 【LeetCode】Subsets 解题报告 。

题意：给定了数组，数组中的数有重复，要求出所有子集。

思路还是回溯。我沿着上道题的思路，分别回溯求长度为0~n的子集。并且在回溯过程中判断重复的子集。

public class Solution {
    List<List<Integer>> ans = new ArrayList<List<Integer>>();
    int[] num = null;

    public List<List<Integer>> subsetsWithDup(int[] num) {
        this.num = num;

        // Sort first!
        Arrays.sort(num);

        // Add the subsets of length from 0 to num.length by back tracking.
        for (int i = 0; i <= num.length; i++) {
            backTracking(new ArrayList<Integer>(), 0, i);
        }

        return ans;
    }

    public void backTracking(List<Integer> cur, int from, int cnt) {
        if (cur.size() == cnt) {
            // Find if cur is in the result.
            boolean hasSame = false;
            int k = ans.size() - 1;
            while (k >= 0 && ans.get(k).size() == cnt) {
                List<Integer> last = ans.get(k);
                int i = 0;
                for ( ; i < cnt; i++) {
                    if (last.get(i) != cur.get(i)) {
                        break;
                    }
                }
                if (i == cnt) {
                    hasSame = true;
                    break;
                }
                k--;
            }

            // If cur is not the same as last subset, add it to the result set.
            if (!hasSame) {
                List<Integer> list = new ArrayList<Integer>(cur);
                ans.add(list);
            }
        } else {
            // Go on back tracking.
            for (int i = from; i < num.length; i++) {
                cur.add(num[i]);
                backTracking(cur, i + 1, cnt);
                cur.remove(new Integer(num[i]));
            }
        }
    }
}