1. Two Sum

🍏🍏🍏

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

You may assume that each input would have exactly one solution, and you may not use the same element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

Solution 1:

class Solution {
    public int[] twoSum(int[] nums, int target) {
        int[] resultArray = new int[2];
        for (int i = 0; i < nums.length; i++) {
            for (int j = i + 1; j < nums.length; j++) {
                if (nums[i] + nums[j] == target) {
                    resultArray[0] = i;
                    resultArray[1] = j;
                    return resultArray;
                }
            }
        }
        return resultArray;
    }
}

Performance: 27ms

Solution 2:

public class Solution {
    public int[] twoSum(int[] nums, int target) {
        if (nums == null){
            return null;
        }
        int[] result = new int[2];
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        for (int i = 0; i < nums.length; i++){
            if (map.containsKey(target - nums[i])){
                result[0] = map.get(target - nums[i]);
                result[1] = i;
                return result;
            } else {
                map.put(nums[i], i);
            }
        }
        return result;
    }
}

Performance: 9ms

2. Add Two Numbers

🍋🍋🍋

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.

You may assume the two numbers do not contain any leading zero, except the number 0 itself.

Example:

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8
Explanation: 342 + 465 = 807.

Solution 1:

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        //Input Validation
        if (l1 == null && l2 == null){
            return null;
        }
        if (isNotValidListNode(l1)) {
            return l2;
        }
        if (isNotValidListNode(l2)) {
            return l1;
        }
        //Create dummyListNode head to record the head node address.
        ListNode dummyListNode = new ListNode(1);
        ListNode currentNode = dummyListNode;
        int sum = 0;
        int carry = 0;
        while (l1 != null && l2 != null){
            sum = l1.val + l2.val + carry;
            currentNode.next = new ListNode(sum % 10);
            carry = sum >= 10 ? 1 : 0;

            l1 = l1.next;
            l2 = l2.next;
            currentNode = currentNode.next;
        }
        while (l1 != null){
            sum = l1.val + carry;
            currentNode.next = new ListNode(sum % 10);
            carry = sum >= 10 ? 1 : 0;
            l1 = l1.next;
            currentNode = currentNode.next;
        }
        while (l2 != null){
            sum = l2.val + carry;
            currentNode.next = new ListNode(sum % 10);
            carry = sum >= 10 ? 1 : 0;
            l2 = l2.next;
            currentNode = currentNode.next;
        }
        if (carry != 0) {
            currentNode.next = new ListNode(1);
            currentNode = currentNode.next;
        }
        return dummyListNode.next;
    }
    private boolean isNotValidListNode(ListNode node){
        return node == null || (node.val == 0 && node.next == null);
    }
}

Performance: 28ms

Solution 2:

class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode dummy = new ListNode(0);
        ListNode t1 = l1, t2 = l2;
        ListNode cur = dummy;
        int sum = 0;
        while (t1 != null || t2 != null) {
            if (t1 != null) {
                sum += t1.val;
                t1 = t1.next;
            }
            if (t2 != null) {
                sum += t2.val;
                t2 = t2.next;
            }
            cur.next = new ListNode(sum % 10);
            cur = cur.next;
            sum /= 10;
        }
        if (sum == 1) {
            cur.next = new ListNode(1);
        }
        return dummy.next;
    }
}

Performance: 34ms

3. Longest Substring Without Repeating Characters

🍋🍋🍋

Given a string, find the length of the longest substring without repeating characters.

Example 1:

Input: "abcabcbb"
Output: 3
Explanation: The answer is "abc", which the length is 3.

Example 2:

Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.

Example 3:

Input: "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3.
             Note that the answer must be a substring, "pwke" is a subsequence and not a substring.

Solution 1:

class Solution {
    public int lengthOfLongestSubstring(String s) {
        if (s == null || s.length() ==0){
            return 0;
        }
        int length = 1;
        for (int i = 0; i < s.length(); i++) {
            for (int j = i + 1; j <= s.length(); j++) {
                if ((j - i > length) && isValidString(s.substring(i, j))) {
                    length = j - i;
                }
            }
        }
        return length;
    }
    private boolean isValidString(String s) {
        char[] charArray = s.toCharArray();
        HashSet<Character> set = new HashSet<>();
        for (int i = 0; i < charArray.length; i++){
            if (set.contains(charArray[i])){
                return false;
            }
            set.add(charArray[i]);
        }
        return true;
    }
}

Performance: Time Limit Exceeded

Solution 2:

public int lengthOfLongestSubstring(String s) {
    int n = s.length(), ans = 0;
    Map<Character, Integer> map = new HashMap<>(); // current index of character
    // try to extend the range [i, j]
    for (int j = 0, i = 0; j < n; j++) {
        if (map.containsKey(s.charAt(j))) {
            i = Math.max(map.get(s.charAt(j)), i);
        }
        ans = Math.max(ans, j - i + 1);
        map.put(s.charAt(j), j + 1);
    }
    return ans;
}

Performance: 66ms

4. Median of Two Sorted Arrays

🌶🌶🌶

There are two sorted arrays nums1 and nums2 of size m and n respectively.

Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).

You may assume nums1 and nums2 cannot be both empty.

Example 1:

nums1 = [1, 3]
nums2 = [2]

The median is 2.0

Example 2:

nums1 = [1, 2]
nums2 = [3, 4]

The median is (2 + 3)/2 = 2.5

class Solution {
    public double findMedianSortedArrays(int[] A, int[] B) {
        int m = A.length;
        int n = B.length;
        if (m > n) { // to ensure m<=n
            int[] temp = A; A = B; B = temp;
            int tmp = m; m = n; n = tmp;
        }
        int iMin = 0, iMax = m, halfLen = (m + n + 1) / 2;
        while (iMin <= iMax) {
            int i = (iMin + iMax) / 2;
            int j = halfLen - i;
            if (i < iMax && B[j-1] > A[i]){
                iMin = i + 1; // i is too small
            }
            else if (i > iMin && A[i-1] > B[j]) {
                iMax = i - 1; // i is too big
            }
            else { // i is perfect
                int maxLeft = 0;
                if (i == 0) { maxLeft = B[j-1]; }
                else if (j == 0) { maxLeft = A[i-1]; }
                else { maxLeft = Math.max(A[i-1], B[j-1]); }
                if ( (m + n) % 2 == 1 ) { return maxLeft; }

                int minRight = 0;
                if (i == m) { minRight = B[j]; }
                else if (j == n) { minRight = A[i]; }
                else { minRight = Math.min(B[j], A[i]); }

                return (maxLeft + minRight) / 2.0;
            }
        }
        return 0.0;
    }
}