backtracking问题

backtracking最基础的问题是Subsets，即给定一个数组，要求返回其所有子集。

Given a set of distinct integers, nums, return all possible subsets.

Note:

• Elements in a subset must be in non-descending order.
• The solution set must not contain duplicate subsets.

backtracking的原理是深度优先遍历

public class Solution {
    public List<List<Integer>> subsets(int[] nums) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        if (nums == null || nums.length == 0) {
            return result;
        }
        Arrays.sort(nums);
        List<Integer> list = new ArrayList<Integer>();
        subsetsHelper(nums, result, list, 0);
        return result;
    }
    public void subsetsHelper(int[] nums, List<List<Integer>> result, List<Integer> list, int position) {
        result.add(new ArrayList<Integer>(list));
        for (int i = position; i < nums.length; i++) {
            list.add(nums[i]);
            subsetsHelper(nums, result, list, i + 1);
            list.remove(list.size() - 1);
        }
    }
}

 Subsets II

Given a collection of integers that might contain duplicates, nums, return all possible subsets.

Note:

• Elements in a subset must be in non-descending order.
• The solution set must not contain duplicate subsets.

此问题跟Subsets的不同之处在于数组中有重复元素，所以在每次for循环中需要考虑剔除掉重复元素。对subsetsHelper函数加一个判断即可。

    public void subsetsHelper(int[] nums, List<List<Integer>> result, List<Integer> list, int position) {
        result.add(new ArrayList<Integer>(list));
        for (int i = position; i < nums.length; i++) {
            if (i != position && nums[i] == nums[i - 1]) {
                continue;
            }
            list.add(nums[i]);
            subsetsHelper(nums, result, list, i + 1);
            list.remove(list.size() - 1);
        }
    }