93.复原IP地址

题目
文章讲解
视频讲解

思路：每轮开始的位置需要变化就需要设置start

class Solution {
    List<String> result = new ArrayList<>();

    public List<String> restoreIpAddresses(String s) {
        if (s.length() < 4 ||s.length() > 12)
            return result;
        backTrack(s, 0, 0);
        return result;
    }

    private void backTrack(String s, int startIndex, int pointNum) {
        if (pointNum == 3) {
            if (isValid(s, startIndex, s.length() - 1)) {
                result.add(s);
            }
            return;
        }

        for (int i = startIndex; i < s.length(); i++) {
            if (isValid(s, startIndex, i)) {
                s = s.substring(0, i + 1) + "." + s.substring(i + 1);
                pointNum++;
                backTrack(s, i + 2, pointNum);
                pointNum--;
                s = s.substring(0, i + 1) + s.substring(i + 2);
            } else
                break;
        }
    }

    private boolean isValid(String s, int start, int end) {
        if (start > end)
            return false;
        if (s.charAt(start) == '0' && start != end)
            return false;
        int num = 0;
        for (int i = start; i <= end; i++) {
            if (s.charAt(i) > '9' || s.charAt(i) < '0')
                return false;
            num = num * 10 + (s.charAt(i) - '0');
            if (num > 255)
                return false;
        }
        return true;
    }
}

78.子集

题目
文章讲解
视频讲解

思路：

class Solution {
    List<List<Integer>> result = new ArrayList<>();
    List<Integer> path = new LinkedList<>();

    public List<List<Integer>> subsets(int[] nums) {
        backTracing(nums, 0);
        return result;
    }

    private void backTracing(int[] nums, int startIndex) {
        result.add(new ArrayList(path));//第一遍遍历为空集
        if (startIndex >= nums.length) {
            return;
        }

        for (int i = startIndex; i < nums.length; i++) {
            path.add(nums[i]);
            backTracing(nums, i + 1);
            path.removeLast();
        }
    }
}

90.子集II

题目
文章讲解
视频讲解

思路：利用used数组进行去重
问题：nums[i] == nums[i - 1]去重为什么不用nums[i] == nums[i +1] 解答，编程中的 i 所指都是当前的，要与前一个比较，当前与之前属于已知，后面的是未知。

class Solution {
    List<List<Integer>> result = new ArrayList<>();
    LinkedList<Integer> path = new LinkedList<>();
    boolean[] used;

    public List<List<Integer>> subsetsWithDup(int[] nums) {
        if (nums.length == 0) {
            result.add(path);
            return result;
        }
        Arrays.sort(nums);
        used = new boolean[nums.length];
        backTracing(nums, 0);
        return result;
    }

    private void backTracing(int[] nums, int startIndex) {
        result.add(new ArrayList<>(path));
        if (startIndex >= nums.length)
            return;
        for (int i = startIndex; i < nums.length; i++) {
            if (i > 0 && nums[i] == nums[i - 1] && !used[i - 1]) {
                continue;
            }
            used[i] = true;
            path.add(nums[i]);
            backTracing(nums, i + 1);
            path.removeLast();
            used[i] = false;
        }
    }
}