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].

算法：简单的穷举法即可

代码：

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

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.

Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
Output: 7 -> 0 -> 8

算法：注意考虑加法的进位，然后按照常规方法即可

代码：

    public static ListNode addTwoNumbers(ListNode l1, ListNode l2) {
        if (l1 == null) {
            return l2;
        } else if (l2 == null) {
            return l1;
        }
        ListNode result = null;
        ListNode end = null;
        int cal = 0;
        while (l1 != null && l2 != null) {
            int value = l1.val + l2.val + cal;
            cal = value / 10;
            value = value % 10;
            ListNode newNode = new ListNode(value);
            if (result == null) {
                result = end = newNode;
            } else {
                end.next = newNode;
                end = newNode;
            }
            l1 = l1.next;
            l2 = l2.next;
        }
        while (l1 != null) {
            int value = l1.val + cal;
            cal = value / 10;
            value = value % 10;
            ListNode newNode = new ListNode(value);
            end.next = newNode;
            end = newNode;
            l1 = l1.next;

        }
        while (l2 != null) {
            int value = l2.val + cal;
            cal = value / 10;
            value = value % 10;
            ListNode newNode = new ListNode(value);
            end.next = newNode;
            end = newNode;
            l2 = l2.next;
        }
        while (cal != 0) {
            int value = cal;
            cal = value / 10;
            value = value % 10;
            ListNode newNode = new ListNode(value);
            end.next = newNode;
            end = newNode;
        }
        return result;
    }

3. Longest Substring Without Repeating Characters

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

Examples:

Given "abcabcbb", the answer is "abc", which the length is 3.

Given "bbbbb", the answer is "b", with the length of 1.

Given "pwwkew", 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.

算法：遍历整个字符串，如果遇到重复字符，则计算之前的子串长度，并更新最大值

代码：

    public static int lengthOfLongestSubstring(String s){
        int max = 0;
        int []charMap = new int[256];
        int start = -1;
        for (int i = 0; i < s.length(); i++) {
            if (charMap[s.charAt(i) + 127] - 1 > start)
                start = charMap[s.charAt(i) + 127] - 1;
            charMap[s.charAt(i) + 127] = i + 1;
            max = max > i - start ? max : i - start;
        }
        return max;
    }