wayetan

/**
 * Given an input string, reverse the string word by word.
 * For example,
 * Given s = "the sky is blue",
 * return "blue is sky the".
 * Clarification:
 * What constitutes a word?
 ***A sequence of non-space characters constitutes a word.
 * Could the input string contain leading or trailing spaces?
 ***Yes. However, your reversed string should not contain leading or trailing spaces.

wayetan

/**
 * Divide two integers without using multiplication, division and mod operator.
 */

public class Solution {
    public int divide(int dividend, int divisor) {
        if(dividend == Integer.MIN_VALUE && divisor == -1) return Integer.MAX_VALUE;
        int res = 0;
        boolean isNeg = (dividend < 0 && divisor > 0) || (dividend > 0 && divisor < 0);
        // a / b

wayetan

/**
 * You are given a string, S, and a list of words, L, that are all of the same length. 
 * Find all starting indices of substring(s) in S that is a concatenation of each word in L exactly once and without any intervening characters.
 * For example, given:
 * S: "barfoothefoobarman"
 * L: ["foo", "bar"]
 * You should return the indices: [0,9].
 * (order does not matter).
 */
public class Solution {

wayetan

/**
 * Given two numbers represented as strings, return multiplication of the numbers as a string.
 * Note: The numbers can be arbitrarily large and are non-negative.
 */
 
public class Solution {
    public String multiply(String num1, String num2) {
        if(num1.equals("0") || num2.equals("0"))
            return "0";
        int l1 = num1.length(), l2 = num2.length();

wayetan

/**
 * Implement regular expression matching with support for '.' and '*'.
 * '.' Matches any single character.
 * '*' Matches zero or more of the preceding element.
 * The matching should cover the entire input string (not partial).
 * The function prototype should be:
 * bool isMatch(const char *s, const char *p)
 * Some examples:
 * isMatch("aa","a") → false
 * isMatch("aa","aa") → true

wayetan

/**
 * Implement wildcard pattern matching with support for '?' and '*'.
 * '?' Matches any single character.
 * '*' Matches any sequence of characters (including the empty sequence).
 * The matching should cover the entire input string (not partial).
 * The function prototype should be:
 * bool isMatch(const char *s, const char *p)
 * Some examples:
 * isMatch("aa","a") → false
 * isMatch("aa","aa") → true

wayetan

/**
 * Given a collection of intervals, merge all overlapping intervals.
 * For example,
 * Given [1,3],[2,6],[8,10],[15,18],
 * return [1,6],[8,10],[15,18].
 */

/**
 * Definition for an interval.
 * public class Interval {

wayetan

/**
 * Given an array of words and a length L, format the text such that each line has exactly L characters and is fully (left and right) justified.
 * You should pack your words in a greedy approach; that is, pack as many words as you can in each line. 
 * Pad extra spaces ' ' when necessary so that each line has exactly L characters.
 * Extra spaces between words should be distributed as evenly as possible. 
 * If the number of spaces on a line do not divide evenly between words, the empty slots on the left will be assigned more spaces than the slots on the right.
 * For the last line of text, it should be left justified and no extra space is inserted between words.
 * For example,
 * words: ["This", "is", "an", "example", "of", "text", "justification."]
 * L: 16.

wayetan

/**
 * Given an absolute path for a file (Unix-style), simplify it.
 * For example,
 * path = "/home/", => "/home"
 * path = "/a/./b/../../c/", => "/c"
 * Corner Cases:
 *  Did you consider the case where path = "/../"?
 *  In this case, you should return "/".
 *  Another corner case is the path might contain multiple slashes '/' together, such as "/home//foo/".
 *  In this case, you should ignore redundant slashes and return "/home/foo".

wayetan

/**
 * Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).
 * For example,
 * S = "ADOBECODEBANC"
 * T = "ABC"
 * Minimum window is "BANC".
 * Note:
 * If there is no such window in S that covers all characters in T, return the emtpy string "".
 * If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.
 */