Description：
Given an array of integers, return indices of the two numbers such that they add up to a specific target.
给定一个整数数组，返回两个数的索引，使它们相加为一个特定的目标。
You may assume that each input would have exactly one solution, and you may not use the same element twice.
可以假设每个输入都有一个解决方案，但是同一个元素不能使用两次。
Example:
Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].
我的解决方法：
public class Solution {
public int[] twoSum(int[] nums, int target) {
int[] res = new int[2];
for (int i = 0; i < nums.length; i++) {
for (int j = i+1; j < nums.length; j++) {
if (target-nums[i]==nums[j]) {
res[0] = i;
res[1] = j;
break;
}
}
}
return res;
}
}
DS方法：
public class Solution {
public int[] twoSum(int[] numbers, int target) {
HashMap<Integer,Integer> hash = new HashMap<Integer,Integer>();
for(int i = 0; i < numbers.length; i++){
Integer diff = (Integer)(target - numbers[i]);
if(hash.containsKey(diff)){
int toReturn[] = {hash.get(diff)+1, i+1};
return toReturn;
}
hash.put(numbers[i], i);
}
return null;
}
}
