LeetCode 1 - 10

1. Two Sum

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

2. Add Two Numbers

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        ListNode result = new ListNode(0);
        ListNode temp = result;
        int sum = 0;
        while(l1 != null || l2 != null){
            sum/=10;
            if(l1 != null){
                sum += l1.val;
                l1 = l1.next;
            }
            if(l2 != null){
                sum += l2.val;
                l2 = l2.next;
            }
            temp.next = new ListNode(sum%10);
            temp = temp.next;
        }
        if(sum/10 >= 1){
            temp.next = new ListNode(sum/10);
        }
        return result.next;
    }
}

3. Longest Substring Without Repeating Characters

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        if(0 == s.length()) return 0;
        Map<Character, Integer> map = new HashMap<>();
        int max = 0;
        for(int i = 0, j = 0; i<s.length(); i++){
            if(map.containsKey(s.charAt(i))){
                j = Math.max(j, map.get(s.charAt(i))+1);
            }
            map.put(s.charAt(i), i);
            max = Math.max(max, i - j + 1);
        }
        return max;
    }
}

4. Median of Two Sorted Arrays

public class Solution {
    public double findMedianSortedArrays(int[] A, int[] B) {
        int m = A.length, n = B.length;
        int l = (m + n + 1) / 2;
        int r = (m + n + 2) / 2;
        return (getkth(A, 0, B, 0, l) + getkth(A, 0, B, 0, r)) / 2.0;
    }

    public double getkth(int[] A, int aStart, int[] B, int bStart, int k) {
        if (aStart > A.length - 1) return B[bStart + k - 1];            
        if (bStart > B.length - 1) return A[aStart + k - 1];                
        if (k == 1) return Math.min(A[aStart], B[bStart]);
    
        int aMid = Integer.MAX_VALUE, bMid = Integer.MAX_VALUE;
        if (aStart + k/2 - 1 < A.length) aMid = A[aStart + k/2 - 1]; 
        if (bStart + k/2 - 1 < B.length) bMid = B[bStart + k/2 - 1];        
    
        if (aMid < bMid) 
            return getkth(A, aStart + k/2, B, bStart,       k - k/2); 
        else 
            return getkth(A, aStart,       B, bStart + k/2, k - k/2);
    }
}

5. Longest Palindromic Substring

public class Solution {
    private int lo, maxLen;
    
    public String longestPalindrome(String s) {
        if (s.length() < 2){
            return s;
        }
    
        for (int i = 0; i < s.length()-1; i++) {
            extendPalindrome(s, i, i);  //assume odd length, try to extend Palindrome as possible
            extendPalindrome(s, i, i+1); //assume even length.
        }
        return s.substring(lo, lo + maxLen);
    }
    
    private void extendPalindrome(String s, int j, int k) {
        while (j >= 0 && k < s.length() && s.charAt(j) == s.charAt(k)) {
            j--;
            k++;
        }
        if (maxLen < k - j - 1) {
            lo = j + 1;
            maxLen = k - j - 1;
        }
    }
}