Two Sum_LeetCode

  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].

解题思路:
通过题目给的vector,用最基础的二重循环遍历查看每组是否符合条件。
复杂度为O(n^2)

(如果有更好的解题方法会继续更新）

class Solution {
public:
    vector<int> twoSum(vector<int>& nums, int target) {
        vector<int>result;
        for (int i = 0; i < nums.size()-1;i++) {
            for (int j = i+1; j < nums.size();j++) {
                if (nums[i] + nums[j] == target) {
                    result.push_back(i);
                    result.push_back(j);
                    return result;
                }
            }
        }
        return result;
    }
};

一开始做的时候用成length了，这里是vector， 应该用size。

并且给result赋值的时候习惯性用下标直接赋值，出现了

Runtime Error Message:
reference binding to null pointer of type ‘struct value_type‘

的问题。换成push_back赋值就过了。

Two Sum_LeetCode

原文：http://www.cnblogs.com/SYSU-Bango/p/7502226.html