LeetCode #1 Two Sum

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.

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Example

Given nums = [2, 7, 11, 15], target = 9,
Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

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Analysis

题目难度为：easy，实际上解决方法也十分简单，可以单纯使用数组来解决，两次遍历所给数组

for(i=0; i<nums.size(); ++i)    for (j = i+1; j<nums.size(); ++j)        if (nums[i]+nums[j] == target)            return vector<int> res(i,j)

当然，本题中没有使用这种方法，但是思想是一样的，使用了c++11中新提供的unordered_map数据结构来实现，具体代码如下所示。

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Code（c++)

class Solution {public:    vector<int> twoSum(vector<int>& nums, int target) {        unordered_map<int, int> index;        vector<int> output;        for (int i = 0; i < nums.size(); ++i) {            int temp = target - nums[i];            if (index.find(temp) != index.end()) {                output.push_back(index[temp]);                output.push_back(i);                return output;            }            index[nums[i]] = i;        }        return output;    }};