Description：
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 code:
public class Solution {
public int[] twoSum(int[] nums, int target) {
int[] ans = new int[2];
for(int i = 0; i < nums.length-1; i++){
for(int j = i+1; j< nums.length;j++){
if((nums[i]+nums[j])==target){
ans[0] = i;
ans[1] = j;
break;
}
}
}
return ans;
}
}
问题：代码复杂度过高，抽空看一下复杂度为O（n）的代码。