Two Sum(easy)

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].

我的解答：

class Solution {public: vector<int> twoSum(vector<int>& nums, int target) {  vector<int> outcome;  bool found = 0;  int a = 0;  int b = 0;  for (int i = 0; i < nums.size(); i++) {   a = i;   for (int j = i + 1; j < nums.size(); j++) {    if (nums[i] + nums[j] == target) {     b = j;     outcome.push_back(a);     outcome.push_back(b);     found = 1;     break;    }
   }   if (found)    break;  }  return outcome; }};
因为很久没有打代码了，因此开始先选了一道easy类型的，解答过程中果然会碰到不少问题。我的思路非常简单，就是用两个循环一个一个的试，这里面我一开始图方便把大于target的都跳过，后来发现这样会把负数忽略掉，因此不能有任何附加的判断；另外leetcode上好像是不考虑两个重复的，因此第二个循环开始的系数是第一个循环的当前系数加1