算法学习day24-回溯

复原IP地址

核心思路

也是用切割法, 只是切割完之后, 判断一下切割的元素是否符合ip地址的要求: 不以0开头、<=255

java 复制代码

public List<String> restoreIpAddresses(String s) {
    findIp(s, 0, 0, new StringBuilder());
    return res;
}

private final List<String> res = new ArrayList<>();

public void findIp(String s, int index, int count, StringBuilder sb) {
    if (count > 4) {
        return;
    }
    if (count == 4 && index == s.length()) {
        res.add(sb.substring(0, sb.length() - 1));
        return;
    }
    if (count == 4 && index < s.length()) {
        return;
    }
    for (int i = index; i < s.length(); i++) {
        String substring = s.substring(index, i + 1);
        if (isIp(substring)) {
            sb.append(substring);
            sb.append(".");
            count++;
        } else if (substring.length() > 3) {
            break;
        } else {
            continue;
        }
        findIp(s, i + 1, count, sb);
        sb.delete(sb.length() - substring.length() - 1, sb.length());
        count--;
    }
}

private boolean isIp(String substring) {
    if (substring.length() > 3) {
        return false;
    }
    if (substring.charAt(0) == '0' && substring.length() > 1) {
        return false;
    }
    return Integer.parseInt(substring) <= 255;
}

子集

核心思路

和切割类似, 只是终止条件变了

不是必须执行完整个数组才结束
如果说组合问题和分割问题都是找树的子节点, 那么子集就是找树的所有节点

JAVA 复制代码

public List<List<Integer>> subsets(int[] nums) {
    findChild(nums, 0, new ArrayList<>());
    return subsetsRes;
}

private List<List<Integer>> subsetsRes = new ArrayList<>();

public void findChild(int[] nums, int index, List<Integer> list) {
    subsetsRes.add(new ArrayList<>(list));
    for (int i = index; i < nums.length; i++) {
        int cur = nums[i];
        list.add(cur);
        findChild(nums, i + 1, list);
        list.removeLast();
    }
}

子集2

核心思路

就是子集+组合2的used逻辑

记得排序

java 复制代码

private List<List<Integer>> subsetsRes = new ArrayList<>();

public List<List<Integer>> subsetsWithDup(int[] nums) {
    Arrays.sort(nums);
    findChild1(nums, 0, new ArrayList<>(), new int[nums.length]);
    return subsetsRes;
}

public void findChild1(int[] nums, int index, List<Integer> list, int[] used) {
    subsetsRes.add(new ArrayList<>(list));
    for (int i = index; i < nums.length; i++) {
        if (i > 0 && nums[i] == nums[i - 1] && used[i - 1] == 0) {
            continue;
        }
        int cur = nums[i];
        list.add(cur);
        used[i] = 1;
        findChild1(nums, i + 1, list, used);
        list.removeLast();
        used[i] = 0;
    }
}