题目：
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.
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
翻译：
给定一个整形数组和一个整数target，返回2个元素的下标，它们满足相加的和为target。
你可以假定：每次输入都会恰好有一个满足条件的返回结果。
（参考别人的解法）
方法一：
public static int[] twoSum(int[] nums, int target) {
if(nums == null || nums.length == 0) return null;
HashMap<Integer, Integer> hashMap = new HashMap<>();
for(int i = 0; i < nums.length; i++) {
hashMap.put(nums[i], i);
}
int[] result = null;
for(int j = 0; j < nums.length; j++) {
int firstNum = nums[j];
int secondNum = target - firstNum;
if(hashMap.containsKey(secondNum) && hashMap.get(secondNum) != j) {
result = new int[2];
result[0] = j;
result[1] = hashMap.get(secondNum);
break;
}
}
return result;
}
方法二：
public static int[] twoSum2(int[] nums, int target) {
if(nums == null || nums.length == 0) return null;
int[] result = null;
for(int i = 0; i < nums.length; i++) {
for(int j = i+1; j<nums.length; j++) {
if(nums[i] + nums[j] == target) {
result = new int[2];
result[0] = i;
result[1] = j;
break;
}
}
}
return result;
}
