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].

My Solutions:

方法1：Brute Force. Time: O(n^2); Space: O(1)

方法2：Two Pass Hash Table. Time: O(n); Space: O(n)

方法3：One Pass Hash Table. Time: O(n); Space: O(n)

用HashMap储存key=>数字的index，value=>数字本身。如果target - 数字的结果已经在hashmap里，直接返回两个数字的index

public int[] twoSum(int[] nums, int target) {
    Map<Integer, Integer> map = new HashMap<>();
    for (int i = 0; i < nums.length; i++) {
        int complement = target - nums[i];
        if (map.containsKey(complement) {
            return new int[] {map.get(complment), i};
        }
        map.put(nums[i], i);
    }
    throw new IllegalArgumentException("No two sum solution.");
}