Weekly Contest 167

[lihe]

Convert Binary Number in a Linked List to Integer

Given head which is a reference node to a singly-linked list. The value of each node in the linked list is either 0 or 1. The linked list holds the binary representation of a number.

Return the decimal value of the number in the linked list.

Example 1:

Input: head = [1,0,1]
Output: 5
Explanation: (101) in base 2 = (5) in base 10

Example 2:

Input: head = [0]
Output: 0

Example 3:

Input: head = [1]
Output: 1

Example 4:

Input: head = [1,0,0,1,0,0,1,1,1,0,0,0,0,0,0]
Output: 18880

Example 5:

Input: head = [0,0]
Output: 0

Constraints:

• The Linked List is not empty.
• Number of nodes will not exceed 30.
• Each node's value is either 0 or 1.

class Solution {
public:
    int getDecimalValue(ListNode* head) {
        if(head == NULL)
            return 0;
        if(head->next == NULL)
            return head->val;
        int num = 0;
        while (head){
            num *= 2;
            num += head->val;
            head = head -> next;
        }
        return num;
    }
};

Sequential Digits

An integer has sequential digits if and only if each digit in the number is one more than the previous digit.

Return a sorted list of all the integers in the range [low, high] inclusive that have sequential digits.

Example 1:

Input: low = 100, high = 300
Output: [123,234]

Example 2:

Input: low = 1000, high = 13000
Output: [1234,2345,3456,4567,5678,6789,12345]

Constraints:

• 10 <= low <= high <= 10^9

算法思路：一共36个数，通过打表即可解决。

class Solution {
    public List<Integer> sequentialDigits(int low, int high) {
        int[] list = {12, 23, 34, 45, 56, 67, 78, 89,
                123, 234, 345, 456, 567, 678, 789,
        1234, 2345, 3456, 4567, 5678, 6789,
        12345, 23456, 34567, 45678, 56789,
        123456, 234567, 345678, 456789,
        1234567, 2345678, 3456789,
        12345678, 23456789,
        123456789};

        List result = new LinkedList();
        for(int i = 0; i < list.length; i++){
            if(list[i] <= high && list[i] >= low){
                result.add(list[i]);
            }
        }
        return result;
    }
}

Maximum Side Length of a Square with Sum Less than or Equal to Threshold

Given a m x n matrix mat and an integer threshold. Return the maximum side-length of a square with a sum less than or equal to threshold or return 0 if there is no such square.

Example 1:

Input: mat = [[1,1,3,2,4,3,2],[1,1,3,2,4,3,2],[1,1,3,2,4,3,2]], threshold = 4
Output: 2
Explanation: The maximum side length of square with sum less than 4 is 2 as shown.

Example 2:

Input: mat = [[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2],[2,2,2,2,2]], threshold = 1
Output: 0

Example 3:

Input: mat = [[1,1,1,1],[1,0,0,0],[1,0,0,0],[1,0,0,0]], threshold = 6
Output: 3

Example 4:

Input: mat = [[18,70],[61,1],[25,85],[14,40],[11,96],[97,96],[63,45]], threshold = 40184
Output: 2

Constraints:

• 1 <= m, n <= 300
• m == mat.length
• n == mat[i].length
• 0 <= mat[i][j] <= 10000
• 0 <= threshold <= 10^5

class Solution {
    public int maxSideLength(int[][] mat, int threshold) {
        int result = 0;

        for(int i = 0; i < mat.length; i++){
            for(int j = 0; j < mat[0].length; j++){
                if(j != 0){
                    mat[i][j] += mat[i][j - 1];
                }
                int len = 0;
                int maxlen = Math.min(i, j) + 1;
                while (len < maxlen){
                    int area = 0;
                    for(int k = 0; k < len + 1; k++){
                        int prefix = j - len - 1 < 0 ? 0 : mat[i - k][j - len - 1];
                        area += mat[i - k][j] - prefix;
                    }
                    if(area > threshold) {
                        break;
                    }
                    len++;
                }
                result =  len > result ? len : result;
            }
        }
        return result;
    }
}