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【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])); } } } }
【LeetCode】Subsets II 解题报告
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