Two Sum

一直听说LeetCode，今天第一次刷题，之前没有接触过算法，记录自己的刷题。

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]

思路：两个数整数相加等于targe，且数组中都为大于零，故需要查找targe-x的位置

解法：

public int[] twoSum(int[] nums, int target) {
        int[] result = new int[2];
        Map<Integer, Integer> map = new HashMap<Integer, Integer>();
        for (int i = 0; i < nums.length; i++) {
            int temp = nums[i];
            if (map.containsKey(temp)) {
                result[0] = map.get(temp);
                result[1] = i;
                return result;
            }
            //如果没有，就put到map里面
            else {
                map.put(target - nums[i], i);
            }
        }
        throw new IllegalArgumentException("No two sum solution");
    }

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这道题作为第一题还是很简单的，开始想把数组通过asList转为List，通过indexof去查找targe-x的位置。 后来突然想到8 个基本类型是无法作为 asList 的参数的。 所以换成map