Day 28 | 93. Restore IP Addresses | 78. Subsets | 90. Subsets II

原创已于 2023-01-10 03:59:57 修改 · 393 阅读

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于 2023-01-09 13:07:04 首次发布

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本文介绍了LeetCode中涉及数组、字符串、二叉树和链表的一系列算法问题，包括查找、删除、构建等操作，以及使用回溯法解决子集和IP地址恢复的问题。

LeetCode 93. Restore IP Addresses

Question Link

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

    public List<String> restoreIpAddresses(String s) {
        backTracking(0, s, 0);
        return result;
    }

    void backTracking(int index, String s, int pointSum){
        if(pointSum == 3){
            if(isValid(s, index, s.length() - 1))
                result.add(s);
            return;
        }
        for(int i = index; i < s.length(); i++){
            if(isValid(s, index, i)){
                s = s.substring(0, i + 1) + '.' + s.substring(i+1); // add the dot
                pointSum++;
                backTracking(i + 2, s, pointSum); // The position of the next substring is i+2 after inserting a dot.
                pointSum--;
                s = s.substring(0, i + 1) + s.substring(i + 2);     // remove the dot
            }else   
                break;
        }
    }

    Boolean isValid(String s, int start, int end) {
        if(start > end)
            return false;
        if(s.charAt(start) == '0' && start < end)
            return false;
        if(Long.parseLong(s.substring(start, end+1)) <= 255)
            return true;
        else
            return false;
    }
}

The position of the next substring is i+2 after inserting a dot.
Use the index to record the start position of the following split.
Use the pointNum to record the number of dots.

LeetCode 78. Subsets

Question Link

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

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

    void backTracking(int index, int[] nums){
        result.add(new ArrayList<>(path));
        if(index >= nums.length)
            return;
        
        for(int i = index; i < nums.length; i++){
            path.add(nums[i]);
            backTracking(i + 1, nums);
            path.remove(path.size() - 1);
        }
    }
}

We should collect the result before the termination condition.

LeetCode 90. Subsets II

Question Link

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

    public List<List<Integer>> subsetsWithDup(int[] nums) {
        Arrays.sort(nums);
        isUsed = new boolean[nums.length];
        backTracking(0, nums);
        return result;
    }

    void backTracking(int index, int[] nums){
        result.add(new ArrayList<>(path));
        if(index >= nums.length)
            return;
        for(int i = index; i < nums.length; i++){
            if(i > 0 && nums[i] == nums[i-1] && isUsed[i-1] == false)
                continue;
            isUsed[i] = true;
            path.add(nums[i]);
            backTracking(i+1, nums);
            path.remove(path.size() - 1);
            isUsed[i] = false;
        }
    }
}