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 thesame element twice.

Example:

Given nums = [2, 7, 11, 15], target = 9,Because nums[0] + nums[1] = 2 + 7 = 9,return [0, 1].

解题思路：1.最容易想到的，莫过于构建两个循环，一组一组的去验证，此时算法复杂度为O(n​2​​)。

public int[] twoSum(int[] nums, int target) {    for (int i = 0; i < nums.length; i++) {        for (int j = i + 1; j < nums.length; j++) {            if (nums[i] + nums[j) == target {                return new int[] { i, j };            }        }    }    throw new IllegalArgumentException("No two sum solution");}

2.用jave,当我们在表中迭代并插入元素时，我们还会回顾一下，看看当前元素的补充是否已经存在于表中。如果存在，我们已经找到了解决方案，并立即返回。此时算法复杂度为$O(n)$。

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(complement), i };        }        map.put(nums[i], i);    }    throw new IllegalArgumentException("No two sum solution");}