Pretty Sure

LeetCode - Palindrome Partitioning

Question Definition

Given a string s, partition s such that every substring of the partition is a palindrome.

Return all possible palindrome partitioning of s.

LeetCode - Linked List Cycle II

Question Definition

Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

LeetCode - Letter Combinations of a Phone Number

Question Definition

Given a digit string, return all possible letter combinations that the number could represent.

LeetCode - Add and Search Word - Data structure design

Question Definition

Design a data structure that supports the following two operations:

void addWord(word) bool search(word) search(word) can search a literal word or a regular expression string containing only letters a-z or .. A . means it can represent any one letter.

LeetCode - Search a 2D Matrix II

Question Definition

Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties:

• Integers in each row are sorted in ascending from left to right.
• Integers in each column are sorted in ascending from top to bottom. For example,

Consider the following matrix:

[
  [1,   4,  7, 11, 15],
  [2,   5,  8, 12, 19],
  [3,   6,  9, 16, 22],
  [10, 13, 14, 17, 24],
  [18, 21, 23, 26, 30]
]

Given target = 5, return true.

Given target = 20, return false.

Java Solution

public boolean searchMatrix(int[][] matrix, int target) {
    if(matrix.length == 0 || matrix[0].length == 0) return false;
    int i = 0;
    int left = 0;
    int right = matrix[0].length - 1;
    while(true){
        int mid = (left + right) / 2;
        if(matrix[i][mid] == target)
            return true;
        if(matrix[i][mid] < target)
            left = mid + 1;
        else
            right = mid - 1;
        if(left > right){
            if(left == 0)
                return false;
            if(i == matrix.length - 1)
                return false;
            i++;
            left = 0;
            right = matrix[i].length - 1;
        }
    }
}